









Book 











Carpentry Made Easy; 

OR, THE 

SCIENCE AND ART OF FRAMING, 

ON A 

NEW AND IMPROVED SYSTEM. 


WITH SPECIFIC INSTRUCTIONS FOR 

BUILDING BALLOON FRAMES, BARN FRAMES, MILL FRAMES, WARE¬ 
HOUSES, CHURCH SPIRES, ETC. 


COMPRISING ALSO 


A SYSTEM OF BRIDGE BUILDING; 


WITH 


BiXiiiS, ~U!?stt-m-ates of Cost, Valuable Tables. 

Jortti-fow YMatcs ;itul near Fwo UuuuUtrt Jipmi. 


SECOND EDITION, ENLARGED AND IMPROVED. 





BY WILLIAM E. BELL, 

> > 

ARCHITECT AND PRACTICAL BUILDER. 


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PHILADELPHIA: 
















1 



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Entered according to Act of Congress, in the year 1857, by 
WILLIAM E. BELL, 

In the Clerk's Office of the District Court of the United States in and for the Eastern District of Pennsylvania. 


0 








Entered according to Act of Congress, in the year 1875, by 
WILLIAM E. BELL, 
















PREFACE TO THE SECOND EDITION. 


The “ Art and Science of Carpentry ” was first presented to 
the public in 1858. I then made some efforts to introduce it 
into general use, and especially to bring it to the notice of 
mechanics and their apprentices. At first these efforts met 
with only partial success. Then the cry of “hard times” was 
no humbug; many good workmen were out of employment 
and money was really scarce. The public has also been so 
often imposed upon by charlatans and quacks of all kinds that 
it is always difficult to introduce anything, however valuable, 
until the popular mind has had time to be convinced of its real 
utility. It was not long before orders for the work began to 
come in, and the first issue of a thousand copies was exhausted. 
Another thousand was then issued and sold, and the third was 
commenced when the publishers and my friends generally, who 
had become fully convinced of the value of the work, urged me 
to prepare a revised and improved edition. This I have now 
undertaken, and the work is here presented, revised, enlarged, 
and improved in all its departments, in hopes of rendering it 
more intrinsically valuable, as well as more acceptable and 
useful. 

Young men are apt to regard the carpenter’s trade an easy 
one, but there never was a greater mistake, for I believe it from 
experience to be one of the hardest. When I had finished my 
apprenticeship my old master assured me that I had now fully 
learned my trade and was myself a master-workman, and I 
was elated with the thought that I knew it all; but on travel¬ 
ling to other places and working in other shops I was soon 

(iii) 






IV 


PREFACE. 


made to feel my deficiencies, and from that time to this I have 
applied my mind to the study of my business, with what 
success I leave the public to judge. 

In this second edition I have done my utmost to make the 
work so plain that a common school-boy can understand the 
rules and appreciate their value. In contrast with this sim¬ 
plicity, I have now before me a book, just published, which goes 
back to the old way of obtaining the lengths and bevels of dif¬ 
ferent parts of a frame by lines and tangents which I am cer¬ 
tain that not one carpenter in twenty can use understanding^, 
and this confirms me in the importance of the remarks on this 
subject contained in the introductory chapter. Only a few 
days ago, a master builder remarked to me that the rule here 
given for finding the backing on hip rafters was alone worth 
the price of the book, and this is only one among many im¬ 
portant rules herein contained, some of which I judge to be of 
greater value than this. 

I have frequently been asked why there is so much geometry 
introduced into this book ? For an answer to this question, I 
beg every one to read the introductory chapter. The rules of 
carpentry are founded upon geometry, and it is only by a course 
of geometrical reasoning that these rules can be demonstrated, 
and only as much of geometry is here given as is necessary for 
this purpose. 

The work having been stereotyped on its first issue, we now 
insert the new matter at the end, which will add very materially 
to its value. We have also revised the whole work carefully, 
and corrected all errors of any importance, and it is here pre¬ 
sented with the authors thanks for past encouragement and 
appreciation, and his hopes for their renewal and continuation. 

The Author. 


Ottawa, Illinois. 






CONTENTS 

. < ■ «>» >- 


PART I.—Geometry. 


Definitions. 17 

Explanations of Mathematical Symbols... 21 

Definitions of Mathematical Terms. 22 

Axioms.,. 22 

Proposition I. Theorem. 23 

Proposition XXX. Problem... t 39 


PART II.—Carpentry 

Use op the' Square in Obtaining Bevels. 

The square described. 

Pitch of the roof. 

Bevels of Rafters... 

Bevels of upper joints and gable-end studding. 

Bevels of Braces.. 

Balloon Frames. 

The sills (light sills). 

The studs.. 

The plates... 

Raising and plumbing the frame. 

The floor joists. 

Upper joists... 

Rafters. 

Gable-end studs. 

Framing the sills. 

Work sides (of timbers). 

To take timber out of wind... 

Spacing for windows and doors. 

Mortices for the studs... 

The gains (for joints). 

The draw bores..... 

A draw pin... 

Supports for the upper joists. 

Crowning of joists.. 

Bridging of joists. 

Lining, or sheeting balloon frames... 

Barn Frames. 

Size of mortices. 

Braces. 

Pitch of the roof.. 

Purlins.. 

Length of the purlin posts... 

Purlin post brace. 

Purlin post brace mortices. 

Upper end bevel of purlin post braces. 

Mill Frames. . 

Cripple studs... 

Trussed partitions..... 


43 

43 

43 

44 

45 
45 
47 
47 
47 
47 

47 

48 
48 

48 

49 

50 
50 

50 

51 
51 
51 
51 

51 

52 

53 
5.3 
53 
55 

55 

56 
56 

56 

57 

58 
58 
60 
61 
61 
63 


( 5 ) 


















































6 


CONTENTS. 


Scarfing. 

Straps and bolts (in scarfing). 

Scarfing over posts. 

Floors in Brick Buildings. 

Trimmer joists. 

Circular Centres. 

Elliptical Centres. 

Arches. 

Hip Roofs... 

Hip Rafters.. 

Side bevel of hip rafters.<... 

Down bevel of hip rafters. 

Backing of hip rafters. 

Lengths and bevels of the jack rafters. 

Hips and Valleys. 

Trapezoidal Hip Roofs. 

Lengths of the irregular hip rafters... 

Bevels of the irregular hip rafters. 

Backing of hip rafters on trapezoidal and other irregular roofs., 

Length of jack rafters. 

Side bevels of jack rafters on the sides of the frame. 

Side bevels of the jack rafters on the slant end of the frame.... 
Down bevel of the jack rafters on the beveled end of the frame 

Octagonal and Hexagonal Roofs. 

Length of the hip rafters. 

Bevels of the hip rafters. 

Backing of the octagonal hip rafters. 

Length of the jack rafters. 

Width of the Building. 

Roofs of Brick and Stone Buildings. 

Lengths and bevels of the braces. 

Dimensions of timbers for figs. 1 and 2. 

Length of straining beam. 

Church Spires. 

Domes... 


PAG* 

64 

64 

65 

66 
66 

67 

68 

69 

70 
70 

70 

71 

71 

72 

73 
73 

73 

74 

75 

75 

76 

76 

77 

78 
78 

78 

79 
79 
79 
81 
81 
82 
83 

89 

90 


PART III. —Bridge Building. 


Straining-Beam Bridges. 93 

Trestle Bridges. 98 

Arch Truss Bridges . 102 

General Principles of Bridge Building . 107 

Backing of hip rafters demonstrated. Ill 

Bracing between posts that stand battering . 113 

Bevels of the braces. 114 

Height of the girder. 114 

Length of the girder and cap... 114 

The long braces. 114 

How to calculate the length of the braces. 115 

Improved straining-beam bridge. 116 

Bill of timber. 116 

Bill of iron. 117 

Large circular center for heavy arches. 118 

Husk or trestle-work for water-tank. 119 

Height and location of tank. 119 

To estimate the capacity of a tank. 119 

Framing and constructing pile bridges. 122 


PART IV. —Explanation of the Tables. 


Definitions of Terms and Phrases used in this Work. 127 

Table I. Length of Common Rafters. 130 

Table II. Length of Hip Rafters. 133 

Table III. Octagonal Roofs. 139 

Table IV. Length of Braces... 141 

Table V. Weight of Square Iron. 142 

Table VI. Weight of Flatiron. 144 

Table VII. Weight of Round Iron. 146 

Table VIII. Weight and Strength of Timber. 148 

How to Calculate the Strength of Timber. 149 






































































INTRODUCTORY CHAPTER. 

SUMMARY VIEW. 


The Science and the Art of Framing. 

No apology is offered for introducing to the Public a work 
on the Science and Art of Framing. By the Science of Fram¬ 
ing is meant the certain knowledge of it, founded on mathe¬ 
matical principles, and for which the master of it can assign 
intelligent reasons, which he knows to be correct; while the 
Art of Framing is the system of rules serving to facilitate the 
'practice of it, but the reasons for which the workman may or 
may not understand. That Carpentry has its rules of Science 
as well as its rules of Art, no intelligent mechanic can doubt. 
The rules of the Art are taught by the master-workman at 
the bench ; or, more commonly, insensibly acquired by habit 
and imitation. But by whom have the rules of the Science 
been laid down, and where have its principles been intelli¬ 
gibly demonstrated? 

« ! 

Something New. 

It is believed that this is the very first attempt ever made 
to bring the Science of Carpentry, properly so called, within 
the scope of practical mechanics. 


( 7 ) 





3 


CARPENTRY MADE EASY. 





Deficiencies of Former Works on Carpentry, 

Whatever has formerly been published on this subject, that 
can, with any degree of propriety, be classed under the head 
of Science, has been only available by professional Architects 
and Designers, being written in technical language and mathe¬ 
matical signs, accompanied by no adequate definitions or ex¬ 
planations; and are as perfectly unintelligible to working-men 
of ordinary education as Chinese or Choctaw. On the other 
hand, the numerous works upon the Art of Carpentry, de 
signed and published for the use of working-men, are sadly 
deficient in details and practical rules. They seem to take 
it for granted that the student is already familiar with his 
business; they furnish him with drafts and plans to work 
from; they tell him authoritatively that such or such an 
angle is the proper bevel for such a part of the frame; but 
they neither tell him why it is so, nor inform him how to 
begin and go on systematically with framing and erecting a 
building. These works are, in fine, chiefly valuable for their 
plates; and even these it is not always possible to work from 
with confidence and accuracy, because no man can work with 
confidence and accuracy in the dark : and he certainly is in 
the dark who does not understand the reasons on which his 
rules are founded. 







The Author’s Experience. 

These facts and reflections have been impressing themselves 
upon the mind of the Author of this work for twenty year? 
past, while he has been serving the Public as a practical car 
penter. During much of this time it has been his fortune 
to have larg Q jobs on hand, employing many journeymen 



INTRODUCTORY CHAPTER. 


9 


mechanics, who claimed to understand their trade, and de¬ 
manded full wages. But it has been one of the most serious 
and oppressive of his cares, that these journeymen knew so 
little of their business. 

Few Good Carpenters. 

They had, by habit, acquired the use of tools, and could 
perform a job of work after it had been laid out for them; 
but not more than one man in ten could himself lay out a 
frame readily and correctly. 

Why Apprentices do not Learn. 

Now, it is not commonly because apprentices are unwilling 
to learn, or incapable of learning, that this is so, but it is be¬ 
cause they have not the adequate instruction to enable them 
to become master-workmen. Their masters are very natu¬ 
rally desirous to appropriate their services to their own best 
advantage; and that is often apparently gained by keeping 
the apprentice constantly at one branch of his business, in 
which he soon becomes a good hand, and is taught but little 
else; and when his time is his own, and he comes to set up 
business for himself, then he is made to feel his deficiencies. 
Should he have assistants and apprentices in his turn, he 
would be unable to give them proper instruction, even were 
he well disposed to do so—for he can teach them nothing 
more than what he knows himself. 

In this condition, the young mechanic applies to books 
to assist him to conquer the mysteries of his Art; but he has 
not been able hitherto to find a work adapted to his wants. 
He anxiously turns the pages of ponderous quarto and folio 
volumes; he is convinced of the prodigious learning of the 


10 


CARPENTRY MADE EASY. 


authors, but he is not instructed by them. On the one 
hand, their practical directions and rules are too meagre; 
and, on the other hand, their mathematical reasoning is too 
technical to yield our young working-man any real benefit 
or satisfaction. May not these faults be remedied ? Is it 
not possible for instruction to be given, which shall be at 
once simple and practical in detail, and comprehensible and 
demonstrative in mathematical reasoning? 

Design of this Work. 

An attempt has been made, in this little work, to answer 
these questions affirmatively; and thus to supply a positive 
want, and to occupy a new field in the literature of Archi¬ 
tecture. Its design is to give plain and practical rules for 
attaining a rapid proficiency in the Art of Carpentry; and 
also to prove the correctness of these rules by mathematical 
science. 


Importance of Geometry to Carpenters. 

No certain and satisfactory knowledge of framing can be 
gained without a previous acquaintance with the primary 
elements of Arithmetic and Geometry. It is presumed that 
a sufficient knowledge of Arithmetic is possessed by most 
mechanics in this country; but Geometry is not so commonly 
understood. It is not taught in our District Schools, and i? 
looked upon as beyond the capacity of common minds. But 
this is a mistake. To mechanical minds, at least, the ele¬ 
ments of Plane Geometry are so easily taught, that they 
seem to them to be almost self-evident at the first careful 
perusal; and mechanics have deprived themselves of much 


introductory chapter. 11 

pleasure, as well as profit, in not having made themselves 
masters of this science. 

Geometry in this Work. 

Part I. is therefore devoted to so much of the Science of 
Geometry as is essential to the complete demonstration and 
thorough understanding of the Science and Art of Carpentry; 
and it is recommended to all mechanics into whose hands 
this volume may fall, to give their days and nights to a 
careful study of this part of the work. It is true that our 
rules and instructions in Carpentry are so plain and minute, 
that they are available to those who do not care to study 
Geometry at all; but the principles on which those rules are 
founded, and consequently the reasons loliy the rules are as they 
are , cannot, from their very nature, be made plain and in¬ 
telligible to any one except by a course of geometrical rear 
eoning. 


New Rules of Carpentry. 

Part II. comprises the main body of the work, and is de¬ 
voted particularly to the framing of buildings. The rules 
for obtaining the bevels of rafters, joists, braces, &c., as ex¬ 
plained in this part of the work, it is believed, have never 
been published before. That such bevels could be so found 
has been known, for several years past, among master- 
builders ; and, to a limited extent, has by that means been 
made public; but this feature of the work will, no doubt, be 
new and useful to some mechanics who have followed the 
business for years, and will be especially useful to apprentices 
and 3 7 oung journeymen who have not yet completed their 
mechanical education. 


12 


CARPENTRY MADE EASY. 


They are Proved and Explained. 

These rules have been here demonstrated by a new and 
rigid course of geometrical reasoning; so that their correctness 
is placed beyond doubt. The demonstrations are often given 
in foot-notes and in smaller print, so as not to interrupt the 
descriptive portion of the work, nor appall those who are not 
mechanically learned, by an imposing display of scientific 
signs and technical terms. In fact, it has been made a lead¬ 
ing object, in the preparation of this work, to convey correct 
mechanical and scientific principles in simple language, 
stripped as much as possible of all technicalities, and adapted 
to the comprehension of plain working-men. 

Bridge Building. 

Part III. comprises a brief practical treatise on the framing 
and construction of Bridges, with bills of timber and iron 
given in detail, by the use of which intelligent carpenters 
can construct almost any kind of a bridge. This part of the 
work does not, however, make any special claims to new 
discoveries, or to much originality; nor is it intended to 
supersede the use of those works specially devoted to Bridge 
Building; but it is believed it will be found more practi¬ 
cally convenient and simple than some others of more im¬ 
posing bulk and of higher price. 

Valuable Tables. 

Part IY. contains a valuable collection of Tables, showing 
the Lengths of Kafters, Hip Bafters, Braces, &c., and also the 
weights of iron, the strength of timber, &c., &c., which will 
be found of the greatest convenience, not only to common 




INTRODUCTORY CHAPTER. 13 

mechanics but to professional designers, architects, and bridge 
builders. Some of these tables have been compiled from 
reliable sources; but the most important of them have been 
calculated and constructed, at a considerable amount of ex¬ 
pense and labor, expressly for this work. 

Plates and Illustrations. 

Nor has any expense been spared in the preparation of the 
plates and illustrations, which are “ got up" in the highest 
style of the art; and it is hoped, and confidently expected, 
that the work, as a whole, will prove to be satisfactory and 
remunerative equally to the Public and to their 

Humble and obedient servant, 

The Author. 


Tiie Author takes great pleasure in acknowledging 
the eminent services rendered him in the literary and 
scientific portions of this work, by E. N. Jencks, A. M., 
Professor of Mathematics and Natural Sciences; and 
the Public cannot fail to appreciate the value of his 
labors in these departments. 

The inception of the work, its original designs, and 
the entire system, are mine. Whatever is round in it 
purely literary and scientific, I cheerfully attribute to 
his assistance. And believing that the work will 
supply a pressing want, and will be useful both to 
those who are devoted to the Mechanic Arts and to 
Amateurs who have felt the necessity of a faithful 
guide in house-building and other structures, especially 
in new settlements, I can confidently commend it to 
them as supplying this deficiency. 



PART 1. 


















































































































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GEOMETRY. 


<•* 


PLATES I. AND II. 

Definitions. 

1. Mathematics is the science of quantity. 

2. Quantity is any thing which can be measured, increased or diminished. 

3. The fundamental Branches of Mathematics are Arithmetic and 
Geometry.* 

4. Arithmetic is the science of numbers. 

5. Geometry is the science of magnitude. 

6. Magnitude has three dimensions : length, breadth, and thickness. 

7. A line has length without thickness. The extremities of a line are 
called points. A point has no magnitude, but position only. 

8. A straight line is the shortest distance between two points. 

9. A curved line is one which changes its direction at every point. It is 
neither straight nor composed of straight lines. 

Thus in Pig. 1, AB is a straight line. ACDB is a broken line, or one 
©omposed of straight lines ; and APB is a curved line. 

10. The single term line is often used in the sense of straight line ; and the 
single term curve, of curved line. 

11. Two lines are parallel when they are everywhere equally distant. 

Pig. 2. 

12. A surface has length and breadth without height or thickness. 

13. A plane is a surface, in which, if any two of its points be joined by a 
straight line, that line will lie wholly on the surface. 

14. A solid, or body, is that which combines the three dimensions of mag¬ 
nitude, having length, breadth, and thickness. 

15. When two straight lines meet each other, the inclination or opening 

* Algebra is a branch of Mathematics, but can scarcely be regarded as equally fundamental 
with Arithmetic and Geometry. 

2 


( 17 ) 




18 


CARPENTRY MADE EASY 


between them is called an angle ; and this angle is said to be greater or less 
as the lines are more or less opened or inclined. 

The vertex of an angle is the point where its sides meet. Thus, in Fig 
3, A is the vertex, and AB and AC are the sides. 

Angles occupy surfaces ; they are therefore quantities ; and like all other 
quantities are susceptible of addition, subtraction, multiplication, and division. 
Thus, in Fig. 4, the angle DCE is the sum of the two angles DCB and BCE 
A nd the angle DCB is the difference of the two angles BCE and DCE. 

An angle is designated by the letter at the vertex, when there is but one 
angle there, as the angle A in Fig. 3; or otherwise by the three letters BAC 
or CAB, the letter at the vertex being always placed in the middle. 

16. When a line, AB, stands on another'line, CD, Fig. 5, so as not to in¬ 
cline either way, AB is said to be perpendicular to CD, and the angle on 
each side of the perpendicular is called a right angle. 

17. Every angle less than a right angle is called an acute angle , as DCB, 
in Fig. 4 ; and every angle greater than a right angle, as ACD, is called an 
obtuse angle. 

18. A polygon is a portion of a plane terminated on all sides by straight 
lines. 

19. An equilateral polygon has all its sides equal, and an equiangular 
polygon has all its angles equal. 

20. A regular polygon is one which is both equilateral and equiangular. 

21. The polygon of three sides is called a triangle; that of four sides, a 
quadrilateral; one of five sides, a pentagon; one of six, a hexagon; one of 
seven, a heptagon; one of eight, an octagon; one of nine, a nonagon ; one of 
ten, a decagon; one of twelve, a dodecagon; one of fifteen, a pentedecagon; 
and so on, according to the numerals of the Greek language. 

22. An equilateral triangle has its three sides equal: Fig. 6. An isosceles 
to'iangle has two of its sides equal. A scalene triangle has all its sides un¬ 
equal. Figs. 7 and 8. 

23. A right-angled triangle contains one right angle. The side opposite 
the right angle is called the hypotenuse. Fig. 9 : AC, opposite the rio-ht 
angle B, is the hypotenuse. 

24. Quadrilaterals are designated according to their figures, as follows : 

The square has its sides all equal, and its angles all right angles. Fig. 10. 

The rectangle, or oblong square, Fig. 11, has all its angles right angles 

and its opposite sides equal and parallel. 

The parallelogram , Fig. 12, has its opposite sides equal and parallel. 
Every rectangle is a parallelogram, but every parallelogram is not a rectangle. 

The rhombus , or lozenge , has its sides all equal without having its angles 
right angles. Fig. 13. 

The trapezium has none of its sides parallel. Fig. 14. 

The trapezoid has two of its sides parallel. Fig. 15. 


GEOMETRY. 


19 


25. The base of any polygon is the side on which it is supposed to stand. 

26. The altitude of a triangle is the perpendicular let fall upon the base 
from the vertex of the angle opposite the base. Thus, in Fig. 6, AB is the 
altitude of the triangle ACD. 

27. The altitude of a parallelogram , or of a trapezoid, is the perpendicu¬ 
lar which measures the distance between two parallel sides taken as bases. 
Thus, in Fig. 12, AB is the altitude of the parallelogram CD. 

28. A diagonal is a line within a polygon, which joins the vertices of two 
angles not adjacent to each other. Thus, in Fig. 16, AC, AD, and AE are 
diagonals. 

29. The area of a polygon is the measure of its surface. 

30 Equivalent polygons are those which contain equal areas. 

31. Equal polygons are those which coincide with each other in all their 
parts, (Ax. 13.) 

32. Similar polygons have the angles of the one equal to the angles of the 
other, each to each, and the sides about the equal angles proportional. 

33. Homologous sides and homologous angles are those which have like 
positions in similar polygons. 

34. The circumference of a circle is a curved line, every point of which 
is equally distant from a point within, called the centre. 

35. The circle is the surface bounded by the circumference. 

36. A radius of a circle is a straight line drawn from the centre to the 
circumference. (The term radius is a Latin word, the plural of which is 
radii. Thus, we say one radius and two radii.) In the same circle all radii 
are equal; all diameters are also equal, and each diameter is double the 
radius. 

37. A diameter of a circle is a straight line drawn through the centre, and 
terminated on both sides by the circumference. In Fig. 17, CD, CG, anil 
CF are radii, and DE is a diameter. 

38. An arc is a portion of the circumference; as AHB in Fig. 17. 

39. A chord is the straight line which connects the two extremities of an 
arc. AB, Fig. 17. 

40. A segment is a portion of a circle included between an arc and its 
choral; as segment AHB, Fig. 17. 

41. A sector is a portion of the circle included between two radii; as 
sector CGF, Fig. 17. 

42. An inscribed angle is one formed by the intersection of two chords 
upon the circumference. ABD and BDE are inscribed angles. 

43. An inscribed polygon is one which, like EFG in Fig. 18, has all its 
angles in the circumference. The circle is then said to circumscribe such a 
figure. In Fig. 17, the triangle CGF is not an inscribed triangle, since all 
the angles do not lie in the circumference; but, in the same Fig., DBE is an 
inscribed triangle. 


20 


CARPENTRY MADE EASY. 


44. A secant is a line which intersects the circumference in two points, 
and lies partly within and partly without the circle. AB is a secant in 
Fig. IT. 

45. A tangent is a straight line touching the circumference in one point 
only. CD is a tangent—0 is the point of contact. 

46. The circumference of every circle is measured by being supposed to be 
divided into 360 equal parts, called degrees; each degree contains 60 min¬ 
utes, and each minute 60 seconds. Degrees, minutes, and seconds, are desig¬ 
nated respectively by these characters, °, ', "; thus, 45° 15' 30" is read 45 
degrees, 15 minutes, and 30 seconds. 

47. Arcs are measured by the number of degrees which they contain 
Thus, in Fig. 19, the arc AE, which contains 90 degrees, is called a quad¬ 
rant, or the quarter of a circumference, because 90° is one quarter of 360°; 
and the arc ACB, which contains 180°, is a semicircumference. 

48. Every angle is also measured by degrees; these degrees being reck, 
oned on an arc included between its sides, described from the vertex of tho 
angle as a centre. Thus, in Fig. 19, the right angle AOE contains 90°; 
and the angle BOD, which is one half a right angle, is called an angle of 45 
degrees, which is the number it contains. 

49. An ellipse is a curved line drawn around two points within, called 
/oci* (A, B, Fig. 20), in such a manner that if, from any point, C, of the 
curve, two lines be drawn, one to each focus, the sum of these two lines, AC 
and BC, shall be equal to the sum of two other lines drawn to the foci from 
any other point of the curve; as DA and DB, or EA and EB. 

50. The centre of an ellipse is the middle point of the line joining the two 
foci. O in Fig. 20. 

51. The diameter of an ellipse is any straight line passing through the 
centre, and terminated on both sides by the curve. 

52. The transverse axis of an ellipse is its longest diameter, or that one 
which passes through the two foci; as GI. 

53. The conjugate axis of an ellipse is its shortest diameter, or that one 
which is perpendicular to the transverse axis; as DF. 

Note. There are several methods employed to describe an ellipse, but the one which is at once 
the most correct and practicable is by means of the instrument called the Trammel, represented 
in Fig. 21. It consists of two grooved rules, mitered together in the middle, so that each arm is 

perfectly perpendicular to the two adjacent arms, and a rod with a movable pencil at P, and two 

mqvable pins, one at A and the other at B. 

The distance from B to P equals half the transverse axis. 

The distance from A to P equals half the conjugate axis. 

The distance from A to B equals half the distance of either focus from the center. 


* The term focus is a Latin word, of which the plural is foci; thus we say one focus and two 
foci. 




GEOMETRY. 


21 


Explanation of Mathematical Symbols. 

In order to facilitate mathematical calculations, it has long been custom¬ 
ary among civilized nations not only to employ figures to represent numbers, 
but also to employ certain other signs or symbols to represent such opera¬ 
tions, in the combinations of numbers and quantities, as are of most frequent 
occurrence; and, by mutual consent, these symbols have come to be gener¬ 
ally known and employed for this purpose. 

1. The sign of addition is written thus -{-, and is read plus; for example 
2-}-3 is read two plus three, and signifies two added to three. 

2. The sign of subtraction is written thus —, and is read minus;* for ex¬ 
ample, 3—2 is read three minus two, and signifies three less two. 

3. The sign of multiplication is written thus X, and is read multiplied 
by ; thus, 3X2 is read three multiplied by two. 

4. The sign of division is written thus -r-, and is read divided by; thus, 
12-^—3 signifies 12 divided by 3. Division is more commonly indicated, how¬ 
ever, by writing the divisor under the dividend, with a line between them, 
in the form of a fraction; thus, ! 3 2 signifies, as before, 12 divided by three. 

5. The sign of equality is written thus =, and is read equals, or is equal 
to ; for example, 2-f3—5 is read thus, two plus three equals five. 

6. The letters of the alphabet, A, B, C, &c., are used as representatives 
of quantities, the exact dimensions of which may either be known or unknown. 
We can let A, for example, stand for a given line, a given angle, a given 
square, or a given solid. Lines are most commonly represented by the two 
letters placed at their extremities; and angles by their three letters, the letter 
at the vertex being always placed in the middle. 

7. A number placed before a quantity is called a co-efficient; thus, 5AB 
is read five AB, or five times AB, or AB multiplied by five: the sign of mul¬ 
tiplication being understood but not written. 

8. A number placed at the right, and a little above a quantity, is called 
an exponent, and indicates how many times a quantity is taken as a factor; 
thus, 5 2 is read five square; it is equal to 5X5, and signifies that five is to 
be multiplied by five, which equals 25 ; also, 5 3 is read five cube ; it is equal 
to 5X5X5, and signifies that five is to be multiplied by five, and that 
product by five, which equals 125. 

9. This sign */ is used to show that a root is to be extracted. A small 
figure is placed in the bosom of the sign, called the index of the root; thus, 
J/ is the sign of the square root, and is the sign of the cube root, &c. 
When no index is written, that of the square root is understood; thus, 
represents the square root of 4. 


* Plus and minus are Latin words, the former moaning more, and the latter U*s; these words 
like the signs, a^e in common use in all civilized countries. 




22 


CARPENTRY MADE EASY. 


Definitions of Mathematical Terms. 

1. An axiom is a self-evident truth. 

2. A theorem is a statement which requires a demonstration, by reasoning 
from such truths as are either self-evident or previously demonstrated. 

3. A problem is a query to be answered, or an operation to be performed. 

4. The term proposition may be applied either to axioms, theorems, or 
problems. 

5. A corollary is a necessary inference drawn from one or more preceding 
propositions. 

6. A scholium is an explanatory remark on one or more preceding propo¬ 
sitions. 

7. An hypothesis is a supposition employed either in the statement or the 
demonstration of a proposition. 

8. The term ratio is employed to denote the quotient arising from dividing 
one number or quantity by another: for example, the ratio of 3 to 12 is 12 di¬ 
vided by 3 ; or; y 2 or 4. The ratio can always be expressed in the form of a 
fraction, whether the divisor is contained in the dividend an exact number of 
times or not; thus the ratio of 2 to 1 is l, the ratio of 5 to 6 is j? ; and so also 

B y 

the ratio of A to B is —, and the ratio of x to y is —. „ 

A x 

9. Proportion is an equality of ratios or an equality of quotients. Thus 
when the quotient arising from dividing one quantity by another is equal to 
the quotient arising from dividing a third quantity by a fourth, then the 
four quantities are said to be in proportion to each other. For example, the 
quotient of 4 divided by 2 equals the quotient of 10 divided by 5 ; or <1 = b° ; 
then these four numbers 2, 5, 4 and 10 are in proportion. 

Proportion is usually indicated by writing the four quantities thus: 
2 : 4 :: 5 : 10, and is read 2 is to 4 as 5 is to 10 ; that is, 2 is just such a part 

of 4 as 5 is of 10 ; for 2 is half of 4, and 5 is half of 10. So also if ? = ? 

A C' 

then we have the proportion A : B :: C : D. 

10. The four quantities of a proportion are called its terms. The first and 
last are called the extremes , and the two middle ones the means of a propor¬ 
tion. The first and third terms are called the antecedents , and the second 
and fourth terms are called the consequents of a proportion. 

Axioms. 

1. A whole quantity is greater than any of its parts. 

2. A whole quantity is equal to the sum of all its parts. 

3. When equals are added to equals, their sums are equal. 

4. When equals are added to unequals, their sums are unequal. 

5. When equals are subtracted from equals, their remainders are equal 


GEOMETRY. 


23 


6. When equals are subtracted from unequals, their remainders are unequal. 

7. When equals are multiplied by equals, their products are equal. 

8. AY hen equals are divided by equals, their quotients are equal. 

9. AY hen two quantities have, each, the same proportion to a third quantity 
they are equal to each other. 

10. All right angles are equal. 

11. AVheu a straight line is perpendicular to one of two parallels it is per¬ 
pendicular to the other also. 

12. Only one straight line can be drawn from one point to another. 

13. T wo magnitudes are equal, when, on being applied to each other, they 
coincide throughout their whole extent. 


Proposition I. Theorem. 


Jf four quantities are in proportion, the product of the two means will equal 
the product of the tioo extremes. 

Numerically. Generally. 

Let 2 : 4 : : 5 : 10 ; A:B::C:D; 

then will 4X5=2X10. BXC=AXD. 

For, since the given quantities are in proportion, their ratios are equal 
(Def. of Terms, 9.) 


And we have, 


4_10 

2“ 5’ 


A - C* 


Multiply both quantities by the divisor of the first ratio, and the quantities 
will still be equal (Ax. 7) ; we shall then have, 


or, 


1=2X™ 


4 B A D 
AX-—AX C ; 

B=AX*t 


Again, multiply both quantities by the divisor of the second ratio, and the de 
sired result is obtained ; namely, 

4X5=2X10. BXC=AXL. 


Proposition II. Theorem. 


When the product of two quantities equals the product of two other quanti - 


ties, then 

two of them are the means, 

and the other two the extremes of a pro - 

portion. 

Numerically. 

Generally. 

Let 

4X5=2X10; 

BXC=AXD; 

then will 

2 : 4 : : 5:10; 

A : B : : C : D ; 

for, divide both the given quantities by one of the factors of the first quantity, 


which will not alter their equality (Ax. 8), and we have, 


4 


2X10 

__ • 


B= 


AXD 

C 




i 


24 CARPENTRY MADE EASI. 

again, divide both quantities by one of the factors of the second quantity, and 
we have, 

4 _10 B_D 

2“ 5* A - C* 

Here we have an equality of ratios, and, by Def. 9, the four quantities are in 
proportion; hence, 

2 : 4 : : 5 : 10. A : B : : C : D. 

Scholium. Quantities are said to be in proportion by inversion, when the 
proportion is read backward ; 

thus, 4 : 2 : : 10 : 5 ; B : A : : D : C ; 

or, 10 : 5 : : 4 : 2. D:C::B:A. 

Quantities are said to be in proportion by alternation , when they are read 
alternately; 

thus, 2 : 5 : : 4 : 10 ; A : C : : B : D ; 

or, 4 : 10 : : 2 : 5. B : D : : A : C. 

Quantities are said to be in proportion by composition, when the sum of the 
antecedents or consequents is compared with either antecedent or consequent, 
thus, 2 + 5 : 5 : : 4 + 10 : 10; A + B : B : : C+D : D; 

or, 2 + 5 : 2 : : 4 + 10 : 4. A+B : A : : C+D : C. 


Proposition III. Theorem. 


When four quantities are in proportion i they will also be in proportion by 
alternation. 


Numerically. 

Let 2 ; 4 : : 5 : 10, 

then will 2 : 5 : : 4 : 10 ; 

for, by Prop. I., 2X10=5X4; 

and, by Prop. II., 2 : 5 : : 4 : 10. 


Generally. 

A : B : : C : D, 
A : C : : B : D ; 
AXD=CXB; 
A : C : : B : D. 


Proposition IV. Theorem. 


When four quantities are in proportion , they will also be in proportion by 
inversion. 


Numerically. 

Let 2 : 4 : : 5 : 10, 

then will 10:5: : 4 : 2 ; 

for, by Prop. I., 10X2=5X4 ; 

and, by Prop. II., 10 : 5 : : 4 : 2. 


Generally. 

A : B : : C : D, 
D : C : : B : A ; 
DXA=CXB; 
D : C : : B : A. 


GEOMETRY. 


25 


Proposition V. Theorem. 

When there are four proportional quantities , and four other proportional 
quantities, having the antecedents the same in both, the consequents will be pro¬ 
portional. 


Numerically. 

Generally. 

Let 2 : 4 : : 5 : 10, and 

A : B :: C : D, 

and 2 : 6 : : 5:15 ; 

A : X :: C : Y ; 

then will 4 : 10 : : 6 : 15, 

Take the first proportion by alternation : 

B : D : : X : Y. 

2 : 5 : : 4 : 10 ; 

A : C:: B :D; 

hence, from equality of ratios (Def.), 


5 10 

C D 

2 4 * 

A - B' 

Take the second proportion by alternation 

• 

• 

2 : 5 : : 6 : 15, 

A : C : : X : Y; 

and, by equality of ratios, we have, 


5 _i5 

C Y 

2 IT 5 

A“X ; 

, 10 15 

D Y 

hence, 

B~X ; 

and from this equality of ratios there results (by Def.), 

4 : 10 : : 6 : 15. 

B : D :: X : Y 

Corollary. When there are two sets of proportional quantities, having an 
antecedent and a consequent of the first equal to an antecedent and a conse* 
quent of the second, the remaining quantities are proportional. 

Proposition VI. 

Theorem. 

When four quantities are in proportion, they are also in proportion by com • 

position. 


Numerically. 

Generally. 

Let 2 : 4 : : 5 : 10, 

A : B : : C : D, 

then will 2+4 : 2 :: 5+10 : 5. 

The first proportion gives (Prop. I.), 

A+B: A::C+D:C. 

2X10=4X5. 

AxD=BxC. 


Add to each of these equal quantities the product of the two antecedents, 
and we have, 

2X10 + 2X5=4X5 + 2X5; AXD + AxC=BxC+A XC; 

or, the same simplified, 

2X10+5=5X4+2; AXD+C=CXB+A; 

hence, by Prop. II., 

2 + 4 : 2 : : 5 + 10 • 5. 


A+B : A :: C+D : C. 


26 


CARPENTRY MADE EASY. 


Proposition VII. Theorem. 


If any two quantities be each multiplied by some other quantify, itoeir 
products will have the same ratio as the quantities themselves. 


Numerically. 

Let 2 and 4 

be any two numbers; 
multiply each by 5 ; 
then 2X5 : 4X5 :: 2 : 4 ; 

for, (2X5)X4=(4X5)X2, 

since the quantities are identical; 
hence, by Prop. II., 2X5 : 4X5 : 


Generally. 

A and 13 


be any two quantities; 



A X S : 13 X S :: A : B ; 
(AXS)XB = (BXS)XA, 


2:4. AxS : BxS : : A : B. 


Proposition VIII. Theorem. 


When two triangles have two sides and the included angle of the one equal to 
two sides and the included angle of the other , each to each, the two triangles are 
equal. 

In the triangles ABC and DEF, let AB = DE, AC=DF, and the angle 
A = angle D ; the triangles themselves will then be equal. For, apply the side 
AB to the equal side DE, so that the point A will fall upon D, and the point 
B upon E ; then since angle A= angle D, the side AC will also fall upon its 
equal side DF, and the point C upon the point F; 
therefore the third side, CB, will fall upon the third side 
FE, and the two triangles will coincide throughout 
their whole extent, and be therefore equal. (Ax. 13.) 

Proposition IX. Theorem. 


c 




When two triangles have two angles and the included 
side of the one, equal to two angles and the included 
side of the other, each to each , the two triangles are 
equal. 

In the triangles ABC and DEF, let the angle 
A— angle D, C = F, and the included side AC = I)F ; then are the triangles 
also equal. 

For, apply the side AC to its equal side DF, placing the point A upon the 
point D, and the point C upon F ; then, since the angle A= angle D, the side 
AB will take the direction of DE, and the point B will fall somewhere upon 
the line DE ; also, since the angle C=angle F, the side CB will take the di¬ 
rection of FE, and the point B will fall somewhere upon the line FE ; and 
since the point B must fall upon both the lines DE and FE, it must fall upon 
E, the only point of coincidence; hence the two triangles coincide through¬ 
out their whole extent, and are therefore equal. (Ax. 13.) 

Cor diary. Every triangle has six parts, namely : three sides and three 




GEOMETRY. 


27 


angles; and whenever two triangles are equal to each other, each of the six 
parts of the one are always equal to the corresponding six parts of the other, 
side to side, and angle to angle. It is to be observed, also, that the equal 
angles are always opposite to the equal sides, and the equal sides opposite the 
equal angles. 



Proposition X. Theorem. 

When a straight line meets another straight line, the sum of the two adjacent 
angles are equal to two right angles. 

Let CD meet AB at D, then is the sum of the two angles ADC and CDB 
equal to two right angles. 

From the point D as a centre, describe the cir¬ 
cumference of a circle, then will the line AB coin¬ 
cide with its diameter, since it passes through the 
centre. (Def.) Angles are measured by the arcs 
intercepted by their sides (Def.) ; and since the 
sides of the angle ADC intercept a portion of the 
semicircumference, AEB, and the sides of the angle 
CDB intercept the remaining portion, then, both 
together intercept a semicircumference, or 180 de¬ 
grees ; but two right angles intercept 180 degrees (Def.) ; therefore the sum of 
the two angles ADC and CDB=two right angles. 

Cor. 1. When one of the given angles is a 
right angle, the other is a right angle also. 

Cor. 2. When one line is perpendicular to 
another, then is the second line also perpendicular to 
the first. 

Let CE be perpendicular to AB, then is AB per¬ 
pendicular to CE. 

For, since CE is perpendicular to AB, both the 
angles ADC and CDB are right angles. Again, since AD is a straight line 
meeting another straight line CE at D, then the sum of ADC-j-ADE=two 
right angles ; but ADC is a right angle ; therefore must ADE be a right 
angle also. Hence AD, or AB, is perpendicular to CE. 

Cor. 3. When any number of angles have their 
vertices at the same point, and lie on the same side 
of a straight line, their sum is equal to two right an 
gies, for they all together intercept an arc of 180 c 


D 


B 


E 








28 


CARPENTKY MADE EASY. 



Proposition XI. Theorem. 

The opposite or vertical angles, formed by the intersection of two straight 
lines, are equal. 

Let AB and CD be two straight lines, intersecting c B 

each other at E, then will AEC = BED. —" 

For the sum of AEC+CEB = two right angles 
(Prop. X.) ; and, for a similar reason, the sum of 
CEB+BED= two right angles. 

Take away from each sum the common angle CEB, and there remains 
AEC=BED. 

In a similar manner it may be proved that CEB=AED. 

Proposition XII. Theorem. 

If tico parallel straight lines meet a third line, the sum of the two interim 
angles, on the same side of the line met, will be equal to two right angles. 

Let the two parallel lines AB and CD meet the 
line EF ; then will BEF + EFG=two right angles. 

Through E draw EG, perpendicular to CD, and 
through F draw FH, parallel with EG. Then, since 
parallels are everywhere equally distant, (Def. 

10), we have EII=GF, and also EG = HF; 
and since AB is perpendicular to EG, it is also perpendicular to HF ; 
(Ax. 11,) and the angles II and G are both right angles; therefore, the two 
triangles, EHF and FGE, are equal, (Prop. VIII.) And, since the angles op¬ 
posite the equal sides are equal, (Prop. IX. Cor.), angle FEII= angle 
EFG. 

But the sum of the angles BEF-j-FEII is equal to two right angles. 
(Prop. X.) Substitute for FEII its equal EFG, and we have BEF-f EFG = 
two right angles. 

Scholium. Where two parallel straight lines meet a third line, the angles 
thus formed take particular names, as follows : 

Interior angles on the same side are those which 
lie within the parallels, and on the same side of the 
secant line. Thus BEF and EFD are interior an¬ 
gles on the same side ; and so also are the angles 
AEF and EFC. 

Alternate angles lie within the parallels, and on op¬ 
posite sides of the secant line, but not adjacent to 
each other. AEF and EFD are alternate angles; 
also, BEF and EFC. 

Alternate exterior angles lie without the parallels, and on opposite sides of 
the secant line. OEB and CFL are alternate exterior angles, and so also 
are AEO and LFD. 









GEOMETRY. 


29 


Opposite exterior and interior angles \ie on the same side of the secant line, 
the one without and the other within the parallels, but not adjacent; thus 
OEB and EFD are opposite exterior and interior angles; so also are BEF 
and DFL. 

Cor. 1. If a straight line meet two parallel lines, the alternate angles will 
be equal. For the sum BEF-f-EFD = two right angles ; also, (by Prop. X), 
BEF-)-AEF= two right angles ; take away from each the angle BEF, and 
there remains EFD=AEF. 

Cor. 2. If a straight line meet two parallel lines, the opposite exterioi 
and interior angles will be equal. For the sum BEF-f EFD = two right angles; 
also, (by Prop. I.), BEF-|-OEB = two right angles; taking from each the 
angle BEF, and there remains EFD=OEB. 

Cor. 3. Hence of the eight angles formed by a line cutting two parallel 
lines obliquely, the four acute angles are equal to each other, and so also are 
the four obtuse angles. 


Proposition XIII. Theorem. 

If two straight lines meet a third line, making the sum of the interior angles 
on the same side equal to two right angles, the two lines will he parallel. 

Let the two lines AB, CD, meet the third line 
EF, so as to make the angles BEF-f EFG equal ah e 

to two right angles; then will AB and CD be pa¬ 
rallel, or everywhere equally distant. 

Through E draw EG perpendicular to CD, and 
trough F draw FH parallel with EG, then the two 
angles FEB-fFEII = two right angles, (by Prop. 

X.); also, the angles FEB-}-EFG= two right angles, by hypothesis; take away 
from each the angle FEB, and there remains the angle FEH-= angle EFG. 
Again, since IIF and EG are parallel by construction, the alternate angles EFII 
and GEF are equal, (by last Prop., Cor. 1) ; hence, the two triangles EFII and 
EFG are equal, (Prop. IX.), having two angles and the included side of the one 
equal to two angles and the included side of the other; and HF, opposite the 
angle FEH, is equal to EG, opposite to its equal angle EFG. (Prop. IX., 
Cor.) But HF and EG measure the distance of the line CD from the line 
AB, at the points H and E respectively. The same demonstration may b* 
applied to any other two points of the line AB; hence the lines AB and 
CD are everywhere equally distant, and therefore parallel. 

Cor. 1. If two straight lines are perpendicular to a third line, they are 
parallel to each other; for the two interior angles on the same side are, in 
that case, both right angles. 

Cor. 2. If a straight line meet two other straight lines, so as to make 
the alternate angles equal to each other, the two lines will be parallel. 






80 


CARPENTRY MADE EASY- 


Let OL meet AB and CD, so as to make AEL = 

EFD ; add to each the angle BEF; we shall then 
have AEL + BEF=EFD+BEF; but AEL-f BEF= A _ 
two right angles (Prop. VIII.) ; hence, EFD-}-BEF= 
two right angles: therefore, AB and CD are parallel. 

Cor. 3. If a straight line, OL, meet two other c 
straight lines, AB and CD, so as to make the ex¬ 
terior angle, OEB, equal to the interior and opposite 
angle, EFD, the two lines will be parallel: for, to each add the angle BEF; 
we shall then have OEB-fBEF=EFD + BEF : but OEB + BEF are equal 
iO two right angles; therefore, EFD + BEF is equal to two right angles; 
and AB and CD are parallel. 

Proposition XIV. Theorem. 

In every parallelogram, the opposite angles are equal. 

Let ABCD be a parallelogram ; then will A = C, 
and B=D. 

Draw the diagonal BD ; then will the triangle 
ADB=the triangle CBD : for the angles ABD and 
BDC are alternate angles and equal (Prop. XII., 

Cor. 1), and the adjacent sides, AB=DC, and BD is common; hence, 
the triangles are equal (Prop. VIII.); therefore the angles A and C, oppo¬ 
site the common side BD, are equal. (Prop. IX., Cor.) In a similar manner 
it may be proved that the angles B and D are equal. 

Cor. 1. The diagonal of a parallelogram divides it into two equal tri¬ 
angles. 

Cor. 2. When two triangles have the three sides of the one equal to the 
three sides of the other, the angles opposite the equal sides are also equal, 
and the triangles themselves are equal. 

Cor. 3. Two parallels, included between two other parallels, are equal. 

Cor. 4. If the opposite sides of a quadrilateral are equal, each to each, 
the equal sides will also be parallel, and the figure will be a parallelogram ; 
for, having drawn the diagonal BD, the triangles ABD and BDC are equal; 
and the angle ADB, opposite AB, is equal to the angle DBC, opposite DC. 
But the two angles, ADB and DBC are alternate angles; therefore, AD is 
parallel with BC. (Prop. XIII., Cor. 2.) ABD and BDC are also equal al¬ 
ternate angles; therefore, AB is parallel with DC, and the figure is a paral¬ 
lelogram. 




Proposition XV. Theorem. 

When two angles have their sides parallel, and lying in the same direction , 
they are equal. 

Let A BC and DEF be two angles, having the side AB, in one, -parallel 







GEOMETRY. 


31 


to DE, in the other, and BC parallel to EE, and lying in the same di¬ 
rection ; then will the two given angles be equal. 

For, produce the side AB till it intersects EF at 
G, then ABC=BGF, for they are opposite in¬ 
terior and exterior angles (Prop., XII. Cor. 2) ; 
also, DEF and BGF are equal for a similar rea¬ 
son : therefore, ABC and DEF, being each equal e 
to BGF, are equal to each other. (Ax. 9.) 



Proposition XVI. Theorem. 

The sum of the three angles of any triangle is equal to two right angles. 

Let ABC be any triangle. Produce the base 
AB to any convenient distance, as D, and draw 
BE parallel with AC ; then will the three 
angles having their vertices at B be equal to the A 
three angles of the given triangle, for the angle B, or ABC, is common ; the axgle 
c=C, for they are alternate angles ; and the angle a= A, for they are opposite 
exterior and interior angles. But the sum of the three angles at B are equal to 
two right angles (Prop. X., Cor. 3) ; hence the sum of the three angles of 
the given triangle, A-f-B-f C, is equal to two right angles. 

Cor. 1. The exterior angle, CBD, of any triangle formed by producing 
the base, is equal to the sum of the two opposite interior angles of the 
triangle. 

Cor. 2. When the sum of two angles of any triangle is known, the third 
angle is found by subtracting that sum from two right angles or 180°. 

Cor. 3. When two angles of one triangle are respectively equal to two 
angles of another triangle, their third angles are also equal, and the triangles 
are equiangular. 

Cor. 4. It is impossible for any triangle to have more than one right 
angle, for if it could have two right angles, the third angle would be nothing. 
Still less can any triangle have more than one obtuse angle. 

Cor. 5. In every right-angled triangle, the sum of the two acute angles is 
equal to one right angle. 



Proposition XVII. Theorem. 

In every isosceles triangle, the angles opposite the equal 
sides are equal. 

In the triangle ABC let AC=BC ; then will angle 
A= angle B. Draw the line CD so as to bisect the angle 
C, that is, so as to divide it into two equal parts; then 
are the two triangles ACD and BCD equal by Prop. 
VIII., having the two angles at C and the two adjacent 
sides equal. Hence, angle A= B. (Prop. IX., Cor.) 


c 








32 


CARPENTRY MARE EASY. 


Cor. 1. Every equilateral triangle is also equiangular. 

Cor. 2. The equality of the triangles ADC, and BDC, proves that the 
line which bisects the vertical angle of an isosceles triangle is perpendicular to 
the base at its middle point, for the two angles at D are each right angles. 
(Prop. X.) 


Proposition XVIII. Theorem. 


When two angles of a triangle are equal , the sides op¬ 
posite them are also equal, and the triangle is isosceles. 

Let the angle A=B, then will the sides AC and BC be 
equal also. 

Draw CD so as to bisect the angle C, then will the two 
triangles be equiangular (Prop. XVI., Cor. 3) ; and the 
side CD being common, the two triangles are equal (Prop. 
IX.); and the side AC, opposite the angle B, is equal to 
the side BC, opposite the equal angle A. 


c 




A 


B 


Proposition XIX. Theorem. 

Parallelograms having equal bases and equal altitudes, contain equal areas, 
or are equivalent. 

Let the two parallelograms, ABCD 
and ABEF, have the same base, AB, 
and the same altitude PS ; then they will 
be equivaleut. 

In the triangles BCE and ADF, the 
sides BC and AD are equal, beiitg opposite sides of the same parallelogram; 
and AF = BE for a similar reason ; the included angle A is equal to the in. 
eluded angle B, since their sides are parallel and lie in the same direction 
(Prop. XV.) ; hence the two triangles are equal. (Prop. VIII.) 

Now, from the whole quadrilateral figure ABCF, take away the triangle 
BCE, and there remains the parallelogram ABEF ; from the same quadri¬ 
lateral take away the equal triangle ADF, and there remains the parallelo¬ 
gram ABCD, which is therefore equivalent to ABEF. 


Proposition XX. Theorem. 

Every triangle contains half the area of a parallelogram of equal base and, 
equal altitude. 

Let ABC be any triangle, and 
ADBE be a parallelogram having 
the same base and altitude ; then will 
the triangle contain half the area of 
the parallelogram. 

Connect C and D, and complete the parallelogram ADCF. The triangleBCP 


E 













GEOMETRY. 


38 


is half the parallelogram FE, (Prop. XIY., Cor. 1) • and the triangle ACF is 
half the parallelogram FD. If from the parallelogram FE we take the paral¬ 
lelogram FD, then the parallelogram AE will remain ; and if from the triangle 
BCF, half the parallelogram FE, we take the triangle ACF, half the parallel 
ogram FD, there will remain the triangle ABC, equal to one half the paral¬ 
lelogram AE. 

Cor. 1. The demonstrations in this and the preceding propositions, are 
equally applicable to rectangles, since every rectangle is also a parallelogram ; 
therefore, every rectangle is equivalent to a parallelogram of the same base 
and altitude. 

Also, every triangle is equivalent to half a rectangle of the same base 
and altitude. 

Cor. 2. Triangles are equivalent to each other, when they have equal bases 
and equal altitudes ; each being half an equivalent parallelogram. 


Proposition XXI. Theorem. 


Two rectangles having the same altitude are proportioned to each other as 
their bases. 


Let the two rectangles AE and CF have equal 
altitudes, then will their surfaces be proportional to 
the length of their bases. 

For, since their altitudes are the same, and their 
angles are all right angles, they may be so applied 
to each other that the whole surface of the shorter 
rectangle shall perfectly coincide with an equal sur¬ 
face of the longer one ; and this coincidence will be perfect as far as there 
a coincidence of their bases, and no further; hence, 

AE : CF : : AB : CD. 



E 

B 


F 


is 


F 


Proposition XXII. Theorem. 

Rectangles are proportioned to each other as the products of their bases mul¬ 
tiplied, by their altitudes. 

Let P be any rectangle, having BC for its base, 
and BF for its altitude; and let N be any other 
rectangle, having AB for its base, and BE for its 
altitude; then, 

P : N : : BCXBF : ABXBE. 

For, place the two rectangles P and X, so that 
the base AB will be the prolongation of the base 
BC, and complete the rectangle M; then, the two rectangles P and M, hav¬ 
ing the same altitude BF, will be proportioned to each other as their bases, 
CB and AB (Prop. XXI.) And, for the same reason, the two rectangles N 


! M 

p 

X 

B C 

E 











34 


CARPENTRY MADE EASY. 


and M, having the same altitude AB, will be to each other as theii bases BE 
and BF ; hence, we have the two proportions : 

P : M : : BC : AB ; and 
M : N : : BF : BE. 

Combining these two proportions, by multiplying the corresponding terms 
together, wm have 

PXM : h T XM : : BCxBF : ABXBE. 

But the quantity M, since it is common to both antecedent and consequent, 
can be omitted ; and the remaining quantities will still be proportional. 
(Prop. VII.) Hence, 

P : 1ST : : BCxBF : ABXBE. 

Cor. 1. Hence the area or surface of any rectangle is measured by the 
product of its base multiplied by its altitude ; and if its base be BC, and its 
altitude BF, its area or measure is BCXBF. 

Cor. 2. Since the sides of every square are all equal, and since all squares 
are rectangles, the area of any square is expressed by the product of a side 
multiplied by itself: so if its side is AB, its area is AB 2 . 

Cor. 3. Since every rectangle is a parallelogram, and since all parallelo¬ 
grams of the same base and altitude are equivalent, (Prop. XIX.), there¬ 
fore the area of any parallelogram is the product of its base by its altitude. 

Cor. 4. Parallelograms of the same base are proportioned to each other 
as their altitudes, and those of the same altitude as their bases; and, in all 
cases, they are proportioned to each other, as the products of their bases by 
their altitudes. 


Proposition XXIII. Theorem. 

The area of any triangle is measured by the product of its base multiplied 
by half its altitude. 

Let ABC be any triangle, of which AB is the 
base, and CD the altitude. This triangle is half the 
parallelogram AE, (Prop. XIV., Cor.) ; but the 
parallelogram is measured by its base, AB, multi¬ 
plied by its altitude, DC ; therefore the triangle is A 
measured by the base multiplied by half the altitude. 

Cor Triangles of the same altitude are proportioned to each other as theii 
bases, and those of the same bases are to each other as their altitudes; and. 
in any case, they are proportioned to each other as the products of their 
bases by their altitudes. 


c E 



Proposition XXIV. Theorem. 

In every right-angled triangle , the square of the hypotenuse is equal to the 
sum of the squares of the other tico sides. 

Let ABC be a triangle, having the angle C a right angle ; then will 
AB 2 =AC 2a -CB 2 . 




GEOMETRY". 


35 


/\ 



•> H 


E 


/ 

F \ 

/ 

I / 

\ 

.• ; 

! / 

\ 

% 

\ j 

i / 

\ ; 

j / 

M \ 1 

i / 

\ i 

/ 
i * 

\ J 

M 

i _ > 


G 


Complete the squares of the three sides of the given triangle, and let M 
represent the square described on AB, rv AB 2 ; let N represent the square 
described on CB, or CB 2 ; and let P represent the square described on AC, 
or AC 2 . Draw the diagonals DB, CE, Cl, and AH, and from C let fall CG 
perpendicular to AB. 

In the two triangles DAB and 
CAE, AC=AD, each being a side of 
the square P ; and AB=AE, each be¬ 
ing a side of the square M; the in¬ 
cluded angle DAB is made up of the 
right angle DAC and the angle CAB ; 
the included angle CAE, in the other 
triangle, is made up of the same angl' 

CAB, and the right angle BAE; 
hence, the angles CAE and DAB ar\ 
equal, and the triangles themselves are 
equal (Prop. YIII.),each having twv 
sides and an included angle equal. 

The triangle DAB is equivalent to 
half the square P, for it has the same 
base AD, and the same altitude AC ; also, the triangle CAE is equivalent 
to half the rectangle FAGE, for it has the same base AE, and the same 
altitude AF; hence, the rectangle FAGE is equivalent to the square P. 
Again, the two triangles ABH rud CBI are equal, having also two sides 
and the included angle of the one equal to two sides and the included angle 
of the other ; and AHB is half of 'he square N, and CBI is half of the rect¬ 
angle FBIG : therefore, the squarr N is equivalent to the rectangle FBIG. 

-But the two rectangles FAGE ard FBIG make up the square M ; hence, 
M=P+N, or AB 3 =AC 2 +CB 2 . 

Cor. 1. In every right-angled tJangle, the square of one side is equiva¬ 
lent to the square of the hypotenus »less the square of the other side. For 
example, in the triangle above, AC 1 «AB 2 —BC 2 ; also, BC 2 =AB 2 —AC 2 . 

Cor. 2. Every square is equal to half the square of its own diagonal. 

Let AD and BC be the diagonals of the square 
ABCD ; through the points A and D draw straight 
lines equal and parallel with BC; and through the 
points B and C draw lines equal and parallel with 
AD ; the figure thus formed will be the square of 
the diagonal CB, or of its equal EF : but this figure 
contains eight equal triangles, of which the given 
equal e contains but four ; hence, 

CB 2 : AB 2 :: 2 : 1 


E 














86 


CARPENTRY MADE EASY. 


and, on extracting the square root of each of the terms of this proportion, wt 
have, 

CB : AB :: v/2 : 1; 

or, the diagonal of a square is proportioned to its side, as the square roo\ 
of two is to one. 


Proposition XXV. Theorem. 

In any triangle, a line drawn parallel to the base divides the other two sides 
proportionally . 

Let ABC be any triangle, and let BE be parallel with 
the base AB. Draw AE and BD. The two triangles 
ADE and CDE, having the same altitude DE, are in 
proportion to each other as their bases AD and CD 
(Prop. XXIII., Cor.) ; also, the two triangles BED and 
CED, having the same altitude ED, are to each other as 
their bases BE and CE ; hence the two proportions: a 

ADE:CDE: :AD:CD; 

BED : CDE : : BE : CE. 

The two triangles ADE and BED are equivalent, having the same base, 
AB, and the same altitude, since the line DE is parallel with BC ; hence, the 
two proportions above having an antecedent and a consequent of one equiva¬ 
lent to an antecedent and a consequent of the other, the remaining terms are 
proportional (Prop. V., Cor.) ; hence, 

AD: CD:: BE: EC; 

and, by composition, AD + CD : CD : : BE + EC : EC ; 
or, AC : CD : : BC : CE ; 

and, by alternation, CD : CE : : AD : BE. 

Proposition XXVI. Theorem. 

In any triangle, the line which bisects the vertical angle, when produced to 
the base, divides the base into two parts, which are proportional to the adja¬ 
cent sides. 

Let ABC be any triangle, and let CE bisect 
the vertical angle C ; then will 

BE : BC : : EA : AC. 

The angles ACE and BCE are equal by hy¬ 
pothesis; draw AD parallel with CE, and pro¬ 
duce it until it intersects the prolongation of BC 
at D ; then will angle D= angle BCE ; for they 

are opposite exterior and interior angles. Also, the angle DAC=ACE, 
since they are alternate angles ; hence, those two angles D and A, in the tri¬ 
angle CAD, equal each other, and the triangle is isosceles. (Prop. XV1JI ) 








GEOMETRY. 37 

In the triangle BAD, since EC is parallel with the base AD, it divides the 
other two sides proportionally (Prop. XXV.), and we have 

BE :BC : :EA : CD; 

but we have proved the triangle CAD to be isosceles; hence, AC = CD. 
Substitute, therefore, in the last proportion, AC for its equal CD, and wo 
have, BE : BC : : EA : AC. 

Proposition XXVII. Theorem. 

All equiangular triangles are similar, and have their homologous sides pro¬ 
portional. 

Let ABC and DEA be two triangles, having 
the angles, C=E, D = CAB, and B=DAE, then 
will their homologous sides be proportional, and 
we shall have 

BA : AD :: BC : AE ; and 
DA : AB :: DE :AC. 

Place the two triangles so that the side AD shall be the prolongation of 
the homologous side AB, and produce DE until it intersects the prolongation 
of BC at F. 

Then since the angles EDA and CAB are equal, the lines FD and CA are 
parallel, for the angles are opposite exterior and interior angles (Prop. XIII. 
Cor.) ; and since the angles DAE and ABC are equal, the lines BF and AE 
are parallel, for those angles are opposite exterior and interior angles also ; 
the figure ACEF is therefore a parallelogram, and has its opposite sides 
equal. 

In the triangle BDF, AC being parallel with the base DF, the other two 
sides are divided proportionally (Prop. XXV.) ; 
and we have 

BA : AD :: BC : CF. But AE=CF ; hence, 

BA : AD :: BC : AE. 

In the same manner it may be proved that, 

DA : AB : : DE : AC. 

Scholium. It is to be observed, that the homologous or proportional side? 
are opposite to the equal angles. 

Cor. Two triangles are similar, and have their homologous sides propor¬ 
tional, when two angles of the one are respectively equal to two angles of the 
other ; for in that ca«e the third angles must also be equal (Prop. XVI: 
Cor.), and the triangles be equiangular. 

Proposition XXVIII. Theorem. 

In even/ convex polygon , the sum of the interior angles is equal to two right 
angles, multiplied by the number of sides of the given polygon, less two. 

Let APCDEF be any convex polygon, and let diagonals be drawn from 





38 


CARPENTRY MADE EASY. 


any one angle, A, to each of the other angles not ad¬ 
jacent to A ; these diagonals will divide the polygon 
into as many triangles, less two, as the polygon has 
sides, whatever the number of the sides may be. 

The sum of the angles of every triangle being equal 
to two right angles (Prop. XVI.), therefore the sum 
of all the angles of the given polygon will be equal to 
twice as many right angles as there are triangles thus formed within it; su 
that, in order to ascertain the entire measure of the angles in any polygon, we 
have only to multiply two right angles by the number of its sides less two. 

Cor. Since 2X2 = 4, the simplest mode of estimating the measure of the 
angles of any polygon, is to multiply the entire number of its sides by two 
right angles, and subtract four from the product. 

A quadrilateral contains four right angles, since 4X2 = 8, and 8—4=4. 

A pentagon contains six right angles, since 5X2 = 10, and 10—4 = 6. 

A hexagon contains eight right angles, since 6X2 = 12, and 12—4=8. 

A heptagon contains ten right angles, since tX2 = 14, and 14—4 = 10. 



Proposition XXIX. Theorem. 

When two triangles have the three sides of the one , respectively parallel or per¬ 
pendicular to the three sides of the other , the two triangles are similar. 

In the two triangles ABC and DEF, 

First. Let the sides be respectively parallel; 
namely, let AB be parallel with DE, BC with EF, 
and AC with DF; then the angles are respectively 
equal; namely, A=D, B=E, and C = F, since 
their sides are respectively parallel and lying in 
the same direction. (Prop. XV.) Hence, their 
homologous sides are proportional, and they are 
similar. (Prop. XXVII.) 

Secondly. Let the sides of the one be respectively 
perpendicular tc the sides of the other ; namely, 

ED perpendicular to AB, FE to BC, and DF to AC ; then they will still be 
equiangular and similar. 

In the quadrilateral LADI, the sum of the four interior angles is equal to 
four right angles (Prop. XXVIII., Cor.) ; but the angles L and I are each 
right angles, since DL is given perpendicular to AC, and ED to AB; there¬ 
fore, the sum of the two angles A and LDI is equal to two right angles; 
but the sum of the angles LDI and LDE equals two right angles (Prop. X.) ; 
take away the common angle LDI from each sum, and there remains, A = 
LDE. 

For similar reasons, B=I)EF, and C=EFD; hence, the two triangles, 
being equiangular, have their homologous sides proportional, and are similar. 
(Prop. XXVII.) 








GEOMETRY. 


39 


Scholium. The homologous sides are those which are perpendicular or 
parallel with each other, since they are also those which lie opposite the eaua 1 
angles. 


Proposition XXX. Problem. 

To inscribe a square within a given circle. 

Let ABCD be the circumference of any circle, 
and let two diameters, AC and BD, be drawn, S / | n. 
intersecting each other at right angles; connect / / | \ 

the ends of these diameters by the chords AB, j / j \ \ 

BC, CD, and DA, then will these chords be bL~~—.- -. 

equal and at right angles with each other, and \ 1 / 

thus form a perfect, inscribed square. \ j / J 

For, AO, BO, DO, and CO are all radii of 'v j / / 

the same circle, and therefore equal (Def.) ; the 

r\ 

four angles at O are right angles by construc¬ 
tion ; hence, the four triangles AOB, BOC, COD, and DOA, are equal (Prop. 

VIII. ), and the chords opposite the equal angles at 0 are also equal. (Prop. 

IX. , Cor.) 

Again, the angles BAD, ADC, DCB, and CBA are all equal, because they 
are each composed of two equal angles ; and, since their sum equals four right 
augles (Prop. XXVIII., Cor.), each one is a right angle, and the figure 
ABCD, having four equal sides and four right angles, is a square. 

Cor. The arcs embraced within the sides of the equal angles at 0, and 
intercepted by the equal chords, are all equal, since each one is the fourth 
part of a circumference, or 90° ; hence, in the same circle, or in equal circles, 
equal chords intercept equal arcs, and equal arcs are intercepted by equal 
chords. 

Proposition XXXI. Problem. 

To inscribe a. regular hexagon and an equilateral triangle within a given 
circle. 

Let ABCDEF be the circumference of any 
circle. Draw the radii AO and BO, in such 
a manner that the chord AB, which connects 
their extremities, shall be equal to the radius 
itself. This chord will be one side of the 
regular, inscribed hexagon. For, the triangle 
ABO, being equilateral, is also equiangular 
(Prop. XVII., Cor.) ; and the sum of its 
three angles, being equal to two right angles 
(Prop. XVI.), each one of its angles is equal 
to tw" thirds of a right angle, or 60°, which 


A 











40 


CARPENTRY MADE EASY. 


is one sixth of a circumference; hence, the sides of the angle AOB intercept 
one sixth of the circumference: therefore, the chord AB, applied six times to 
th.e circumference, will exactly reach around it, and form a regular hexagon; 
for the angles of this hexagon will also be equal, since each one is made up 
of two equal angles, namely, BAO-j-OAF and ABO-f-OBC, &c. 

After having inscribed the regular hexagon, join the vertices of the alter¬ 
nate angles of the hexagon, and the figure thus formed will be an equilateral 
triangle; for its sides are chords which intercept equal arcs, and are there¬ 
fore equal. 



PART I L 













. 



































. 














































































































































m 








































. 





























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CARPENTRY. 




PLATE 3. 

THE USE OF THE SQUARE IN OBTAINING BEVELS. 

Although the square is one of the first instruments placed in the 
hands of the practical carpenter, yet there are many experienced me¬ 
chanics who have never learned all the important uses to which it 
can be applied. And it is claimed as one of the principal merits of 
this work, that it teaches the manner of obtaining the bevels of rafters, 
braces, upper joists, gable-end studding, &c., in the most simple and 
most accurate manner possible, by the use of the square and scratch 
awl alone, without drafts or plans. 

The Square Described. 

The common square is represented in Plate 3, Fig. 1, drawn to the 
scale of one fourth its size. The point 0 is called the corner or the 
heel of the square, the part OH is called the blade, and the part OP 
the tongue. The blade is 24 inches long; the tongue varies in length 
in different squares. We commence at the heel to number the 
inches each way. 


Pitch of the Do of. 

The bevels of rafters, joists, &c., must, of course, vary with the pitch 
of the roof. If the roof is designed to have a quarter pitch, which ia 
the most common inclination for a shingle roof, the peak of the roof 
will be a quarter of the width of the building higher than the top of 
the plates. Although this is called, among builders, a quarter pitch 
yet it would be more simple to call it a half pitch when the roof has 
two sides, which is most commonly the case, for the true inclination 

(43) 



44 


CARPENTRY MADE EASY. 


of each side is 6 inches rise to every foot in width; and in like manner, 
a third pitch is, in reality, a two-thirds inclination to each side of the 
roof, for it has 8 inches rise to every foot in width. 

Bevels of Rafters. 

Let AB, in Fig. 1, represent a rafter which is required to be beveled 
to a quarter pitch. First measure the exact length required, upon the 
edge AB, (whicli will be the upper edge of the rafter when it assumes 
its proper place in the frame,) and let the extreme points A and B be 
marked. Then place the blade of the square upon the point A at the 
12 inch mark, and let the tongue rest upon the edge of the rafter at 
the 6 inch mark; hold the square firmly in this position, and draw 
the line AC along the blade: this line will be the lower end bevel, 
Take the square to the other end of the rafter, and place the 6 inch 
mark on the tongue upon the point B, still having the blade at the 
12 inch mark, and while in this position draw the line DB along the 
tongue: this line will be the upper end bevel required. 

Proceed in a similar manner to mark the bevels for any other pitch, 
placing the 12 inch mark on the blade upon the point A, and that 
mark on the tongue which corresponds with the rise of the roof to 
the foot, on the point B; then the blade of the square will show the 
lower end bevel, and the tongue the upper end bevel. Thus, if the 
roof has a pitch of five inches to the foot, let the square be placed at 12 
and 5; if the roof has 8 inches rise to the foot, place it at 12 and 8, &c. 

The reason of this rule can be explained in few words. In Fig. 2, 
let C represent the middle point of the line which is drawn from the 
top of one plate to the top cf the other; let AB represent a rafter; and 
EC the longest gable-end stud, having its longest edge EC directly 
under the peak of the roof B. The lower end bevel of the rafter rests 
upon the upper surface of the plate, which is horizontal or level, while 
its upper end bevel is perpendicular, resting against the upper end of 
the opposite rafter; so that the upper and lower end bevels of every 
rafter are always at right angles with each other, whatever the pitch 
of the roof may be. r i he tongue of a square is also always at right 
angles with the blade ; and a square can be conceived as having its heel 
at the point C, its blade resting upon the line AC, and its tongue 
standing perpendicularly along the line CB. Now let the distance 
from A to C be supposed to be 1 foot, or 12 inches; then, if the roof is 
designed to rise 6 inches to the foot, the point B will be 6 inches from 
C; T it rises 8 inches to the foot, the point B will be 8 inches from 


USE OF THE SQUARE. 


45 


C, &c.; and in all cases tlie line AC will be one bevel, and the line BC 
the other. 


Bevels of Upper Joists and Gable End Studding. 

The bevel of the upper joists is always the same as the lower end bevel 
of the rafters, and the bevel of ike gable end studding is the same as the 
upper end bevel of the rafters , whatever the pitch of the roof may be. For, 
in respect to the upper joists, it is to be observed that their lower 
surfaces rest upon the plates, and their ends are to be beveled to lit 
the line AB, (Fig. 2), hence the angle BAD, the bevel of the rafter, 
is identical with that of the end of the joist. The bevel of the end 
studding is designed to fit it to the lower surface of the rafter; hence 
the angle DEC is the proper bevel. But DEC=ABC, since they are 
opposite exterior and interior angles. (Part I. Prop. XII., Cor) 

Bevels of Braces. 

Proceed in a similar manner to obtain the bevels of braces. When 
the foot and the head of the brace are to be equally distant from the 
intersection of the two timbers required to be braced, and when the 
angle of their intersection is a right angle, then the brace is said to be 
framed on a regular run , and the bevels will be the same at both ends 
of it, and will always be at an angle of 45°, which is half a right 
angle, or the eighth part of a circle; and this bevel is obtained from 
the square by taking 12 on the blade and 12 on the tongue, or any 
other identical number, the rise being equal to the run* 

But when the foot and the head of the brace are to be at unequal 
distances from the intersection of the timbers, the brace is said to be 
framed on an irregular run , and the bevel at one end will be different 
from that of the other. One rule, however, will answer in all cases. 
First find the length of the brace from the extreme point of one 
shoulder to the extreme point of the other, and mark those points as A 
and B. Then place the blade of the square upon the point A at such 
a distance from the heel as corresponds with the run of the brace, 
while the tongue crosses the edge of the brace at that distance from the 
heel which corresponds with the rise of the brace, and then the blade 
of the square will show one bevel, and the tongue the other. 

For example, a brace is required to be properly beveled for an ir- 


• For the explanation of those terms, rise, run, &c., see the Introduction to the 
Tables. Par' IV. 




46 


CARPENTRY MADE EASY. 


regular run of 4 by 5 feet. Having found tlie length of the brace (by 
Table Ho. 4, or otherwise), and fixed the extreme points of the 
shoulders, then lay on the square at the 4 and the 5 inch marks, and 
describe the bevels along the blade and tongue respectively, as in 
finding the bevels of rafters, 

Fig. 3 represents a small ivory rule, drawn full size. It is intro¬ 
duced here for the purpose of showing the manner of taking the 
measure of hundredths of an inch. It will be perceived that one of the 
inch spaces of the rule is divided into ten parts, by lines running 
down diagonally across ten other horizontal lines. Each of the in¬ 
tersections of these lines measures the hundredth part of an inch ; the 
first line measuring tenths, the second twentieths, &o. 









■ 







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mi • • :%M 












Plate 4. 

FixfJ. 


■H—V 

y 








— 

fi --|~| 


ZL_-~g 

1 

1 

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1 

I 

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I 

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l 

i 

i 




1 






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5 







j 









j 

I 



i 

1 




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2 




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A 

?! 

3"' 

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1 

5. i 

_1 

terrfd 

52 

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Fig. >?, 




Tlii-o.T.'■iiulmi'd* j*Sim. Vi. '.. 












































































































































PLATE 4. 


Balloon Frames. 

As Balloon Frames are the simplest of all, they are the first to claim 
our attention. 


The Sills. 

Where square timber can conveniently be furnished for sills, it is 
best to have it; but small buildings can be very well constructed 
without square sills, even when resting upon blocks only, by using a 
double sot of common joists, with a 2 inch space between them, for 
the tenons of the studding. 

Such a frame, of one story in height, 16 feet long, and 12 feet wide, 
is represented in Plate 4. For this building, joists which are 2 
inches thick and 6 inches wide, will answer. First, for the sills, cut 
two joists 16 feet long, and two others, 11 feet 8 inches long. Spike 
them together at the ends in the form of an oblong square, 16 by 12 
feet, making the outside rim of the sill. 


The Studs. 

Next, frame 13 studs for one side of the building; the two corner 
studs should be 4 inches square, the others 2 by 4. Cut out a 2 inch 
relish, six inches from the foot of each stud, on the face side, leaving 
a tenon on the inside of 6 inches long and 2 inches square, as repre¬ 
sented in Fig. 3. Then cut off the other end of the studs at 10 feet 
from the shoulder. 

The Plates. 

A plate of 2 by 4 stuff, 16 feet long, is now to be nailed flat upon the 
upper ends of the studs, commencing at the front corner, and taking 
care to fix them 14 inches apart, or 16 inches from centre to centre. 
The last space will often be more or less than 14 inches; but it is 
better to have the odd space all at one end, for the convenience of 
the plasterers in lathing. 

Raising and Plumbing the Frame. 

This side of the frame is now ready to be raised. After having 
prepared the other side in the same manner, that can also be raised. 

(47) 


48 


CARPENTRY MATE EASY. 


and the tenons spiked firmly to the inside of the sills. The corners 
should then be plumbed and securely braced. 

The side sills should now be completed by cutting two joists, one 
for each side, each 15 feet 8 inches long, and framing them for the 
support of the floor joists by cutting notches into their upper edges 
2 inches wide and 2J inches deep ; cutting the first notch 16 inches 
from the front end, and the next one just 14 inches from that, and so 
on to the last. After these inside sill-pieces are thus prepared, they 
should be spiked to their places upon the inside of the tenons of the 
studs. 


The Floor Joists. 

The floor joists are to be cut off 11 feet 8 inches long, and their lower 
corners notched off 2 J inches deep ; then they should be fixed in their 
places in the sills, and also spiked to the studs. By this arrangement 
the joists are left one inch higher than the sills for the purpose of 
having the door-sill level with the flooring; the door-sill being 2 
inches, and the flooring 1 inch thick 

Tipper Joists. 

The next thing is to frame the upper joists, the rafters and the 
gable end studs; beveling the ends of each, so as to correspond with 
the pitch of the roof. The bevel is easily found by the use of the 
square, as is explained in Plate 8. The upper joists are equal in 
length to the width of the building. They should be nailed firmly 
upon the top of the plates, the first one being placed 4 inches from 
the end of the plate, to leave room for the end studding. The second 
one should be 14 inches from the first, and the others at the same 
distance from each other, or 16 inches from centre to centre. 

The Rafters. 

The exact length of the rafters is found by the use of Table ISTo. 1, 
Part IV. Look at the left-hand column for the width of the building:, 
and at the top for the rise of the rafter; where those two columns meet 
in the table, the length of the rafter is found in feet, inches, and hun¬ 
dredths of an inch. In this case, the width of the building is 12 feet, 
and the rise of the rafter is 6 inches to the foot, or a quarter pitch; 
therefore the length of the rafter, as given in the table, is (6:8.49; 
6 feet 8 inches and t ‘q 9 g of an inch. The rule for obtaining these lengths 
is of perfect accuracy, and is explained in the introduction to the Tables 


BALLOON FRAMING. 


49 


in such a manner that every carpenter can calculate these lengths for 
himself, from the primary elements, if he chooses. The size of these 
rafters is 2 by 4. 

Gable-End Studs. 

The length of the gable-end studding may be found by first cal¬ 
culating the length of the longest one, which stands under the very 
peak, and then obtaining the lengths of the others from this ; or, by 
first calculating the length of the shortest one next to the corner of 
the building, and then obtaining the lengths of the others from this. 

The length of the middle stud is found by adding to the length of 
the side studding, the rise of the roof and the thickness of the plate ; 
and deducting from that sum the thickness of the rafter, measured on 
the upper end bevel. For example, in this building, the length of the 
side studding from shoulder to shoulder is 10 feet, the rise of the roof is 
3 feet, and the thickness of the plate is 2 inches. These all added are 
18 feet 2 inches, from which deduct 4.47 inches, or 4 \ inches, the thick 
ness of the rafter measured on the upper end bevel, and the result is 
12 feet 10J inches, the length of the middle stud. The next stud, if 
placed 16 inches from this one, from, centre to centre, is 8 inches 
shorter, since the rise of the roof is 6 inches to 12, or 8 inches to 16. 
The next one is 8 inches shorter still, and the others in proportion. 

If it should be thought preferable to commence by first calculating 
the length of the shortest one, it can be done. For example, in this 
building the distance of the inside of the first stud from the outside of 
the building is 20 inches, the rise of the rafter in running 20 inches 
back is 10 inches, to which add 2 inches, thickness of the plate, and 
10 feet for the length of the side studding, and the sum is 11 feet; 
from this deduct the thickness of the rafter, at the upper end bevel, 
4J inches, and the result is 10 feet 7J inches, the length of the shortest 
stud. The length of the next one is found by adding the rise of the 
roof in running the distance, that is, if they are 16 inches apart from 
centre to centre, the difference between them is 8 inches; and so in 
any other pitch, in the proportion of the rise to the run. 

The end studding having been properly beveled and cut off to the 
exact lengths required, they can be raised singly and spiked to the sills 
at the bottom and nailed at the top to the end rafters, and also to the 
upper joists where they intersect them. After the end studs are all 
fixed in their positions, the end sills can finally be completed by 
spiking a joist 11 feet in length to the inside of the studding at each 
end of the frame, 

4 


PLATE 5. 


Plate 5 is designed to represent a balloon frame of a building a 
story and a half high, 16 by 26 feet on the ground, with 12 feet stud¬ 
ding. Two end elevations are given, in order to exhibit different styles 
of roofs. Fig. 2 being a plain roof, of a quarter pitch; and Fig. 3 a 
Gothic roof, the rafters rising 14 inches to the foot. 

Framing the Sills. 

Solid timber, 8 inches square, being furnished for the sills of this 
building, the first business is to frame these. The carpenter will seldom 
have timber furnished to his hand which is perfectly square through¬ 
out its length; by carelessness in hewing, or by the process of season 
ing after being hewed, it will most commonly have become irregular 
and winding. 

Work Sides. 

Having first selected the two best adjoining sides, one for the upper 
side and the other for the front, called work sides , they should be take& 
out of wind in the following manner. 

To take Timber out of Wind. 

Plane off a spot on one of the work sides, a few inches from one end, 
and draw a pencil line square across it; then place the blade of a 
square upon this line, allowing the tongue to hang down as a plummet, 
to keep the blade on its edge. Leave the square in this position, and 
go to the other end of the sill, and place another square upon it, in the 
same manner; then sight across the two squares, and see if they are 
level or parallel with each other. If not, make them so, by cutting 
off* the spots under the squares till they become so; then make the 
other work side square with this one, at these two spots, and draw a 
pencil mark square across both sides: these marks are called 
plumb spots. 

On the upper side of the timber, strike a chalk-line, from one end 
t n the other, at two inches from the front edge; this will be the front 
line for moitices for studs. On this line measure the length of the 
(50) 


Plate 5. 




































































































































































































































BALLOON FRAMING. 51 

sills, and square the ends by it. If the stick is very irregular , it should 
be counter-hewed, and the two work sides made square and straight. 

Spacing for Windows and Doors. 

Next, lay out spaces for windows and doors, leaving a space for the 
doorway 2J or 3 inches more than the width of the door; and leave 
spaces, 7 inches more than the width of the glass, for the windows. 

Mortices for the Studs. 

Then lay out the mortices for the studding, spacing them as de¬ 
scribed in Plate 4. The studding on each side of the doors and win¬ 
dows should be 4 inches* square, as well as those at the corners of the 
building. The rest of the studding may be 2x4. The mortices need 
to be a little more than 2 inches deep, and the tenons 2 inches long. 

The lower joists for this building should be 2X8, and 10 inches 
shorter than the width of the building. They should be placed 16 
inches apart, from centre to centre, as already described. 

The Gains, 

Ae they are called, for receiving the ends of the joists, should be 
cut out of the side sills, 4 inches deep and 2 inches square, and 5 
inches from the front or outside of the sill. Having framed the 
sdls for the studding and joists, they should next be framed for each 
other. Make mortices in the ends of the side sills, 2J inches from the 
upper surface and 2 inches from the end, 2 inches wide an4 5J inches 
long. The inside of the sills should be faced off, along the mortice, 
to within 7J inches of the work side, in front. The length of the side 
sills should be the same, of course, as that of the building; but the 
end sills should be measured from shoulder to shoulder, 15 inches less 
than the width of the building. Make the tenons of these to corres¬ 
pond with the mortices of those which have just been described. 

The Draw Bores. 

The draw bores should be 1 inch in diameter, and 1J inches from 
the face of the mortice. The draw bore through the tenon should be 
, 3 g of an inch nearer the shoulder than that through the mortice, m 
order to draw the work snugly together. 

* 

A Draw Pin. 

The proper way to make a draw pin for an inch bore is, first, to 


52 


CARPENTRY MADE EASY. 


make it an inch square; then cut off the corners, making it eight- 
square, then taper it to a point, the taper extending one third the length 
of the pin. The pin should be about 2 inches longer than the thickness 
ol’ the timber. 

The sills having thus been framed, they can be brought to theii 
places and pinned together, and then the lower joists laid down. 

To Support the Upper Joists. 

This building being a story and a half high, the upper joists are 
laid upon a piece of inch board, from 4 to 6 inches wide, which is let 
into the studs, as seen in the Plate. The bevels and lengths of the 
rafters are found as already described. 

In Fig. 3 the rafters are represented as projecting beyond the 
plate; this projection may be 3 feet, or more, according to each one’s 
fancy; but whatever it may be, it must be added to the length of the 
rafter as given in the table, where it is calculated from the upper and 
outer corner of the plate. The bevel will be the same, whatever the 
additional length may be, as if the rafter did not project at all. In 
this case, the rafter should be cut out to about one half its width, 
where it intersects the plate, and must be spiked securely to the plate. 
The two bevels, at the intersection, will be the same as the upper and 
lower end bevels, and will make a right angle with each other where 
they meet at this place. 

The collar beams can be spiked to the rafters, or they can be dove¬ 
tailed into them. Both methods are represented in the plate. 






















































. 






















Plate 6. 

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PLATE 6. 

Plate No. 6 represents a balloon frame of a two-story building 
18 by 30 feet, with 18 feet studding, to be erected upon a good stone 
or brick wall. Heavy joists, 3 by 10, are used for sills, with the ends 
halved together, and fastened with spikes, as represented in the Plate. 
The lower joists should be 2 by 9 inches, of full length, equal to the 
width of the building. The lower corners are notched off 3 inches, 
and they are spiked to the studding. The mortices for the studs should 
be 1J inches deep, the studding being 2 by 4. The middle joists are 
2 by 9, and arranged as in Plate 5; and the upper, 2 by 7, and ar¬ 
ranged as in Plate 4. 

* 

Crowning of Joists. 

It will almost always happen that one edge of a joist will have be¬ 
come somewhat rounded out, and the other edge rounded in, by the 
process of seasoning; and it is of much importance, especially in long- 
joists of 18 feet or more, to be careful, in placing the joists in a 
building, to place the rounding or crowning side up. 

Bridging of Joists. 

Joists 12 feet long, or over, should also be bridged in one or more 
places, by nailing short pieces of board, 2 or 3 inches wide, in the fo*”- 
of a brace, from the lower edge of one joist to the upper edg v. 
next one; and then another piece, from the lower edge of thi > one r o 
the upper edge of the first one; and so on, throughout the vvhole 
length of the building: having two braces crossing each other a tween 
each joist, beveling the ends so as exactly to fit, which would auu very 
much to the strength of the floor. 

Lining or Sheeting Balloon Frames. 

After an experience of fifteen years in constructing and repairing 
balloon-framed buildings, I have found it best to line the frame on 
the inside for three reasons : 

First — the work is more durable. For, when a frame is lined on 
the outside, (the common way,) it is very difficult to weather-board 
it sufficiently tight, to prevent the rain beating in between the siding 

(53) 


CARPENTRY MADE EASY. 


54 

and the lining, and thus rotting both, since there is so little opportu¬ 
nity there for the moisture to dry out. 

Second —the lining is stiffer and warmer. For, in that case, the lath 
being but half an inch from the lining-boards,* the mortar is pressed 
in between every board, making it almost air-tight. 

Third— the wall itself is made more solid. For the mortar being 
pressed against the lining-boards, is forced both ways, both up and 
down, forming more perfect clinchers. 


* When a building is thus lined on the inside, it is best to lath it in the following 
manner. Single strips of lath are first nailed perpendicularly, sixteen inches apart, 
upon th< lining-boards, and to these the laths for the wall are nailed as usual. 































































































I 

- 












































































































Plate 7. 


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PLATE 7. 


BARN FRAMES. 

Plate 7 represents tlie frame of a barn 30 by 40 feet, and 16 feet 
high between shoulders. 

The sills are 12 inches square; 

Posts and large girders, 10 inches square; 

Plates and girders over main doors, 8 by 10; 

Purlin plates, 6 by 6; 

Purlin posts and small girders, 6 by 8; 

Braces, 4 by 4; and rafters, 2 by 6. 

First proceed to take the timber out of wind, as directed under 
Plate 5. Frame the sills together as represented in the Plate, the 
four short sills being framed into the two long ones, having taken 
care to select the best of the short sills for the ends. 

Size of Mortices. 

The mortices for the end sills should be 3 by 9 inches, with a relish 
of 2J or 3 inches on the outside. The mortices for the middle 
sills may be 3 by 11 inches. The mortices for the corner posts 
should be 3 by 7 inches, and for the middle posts, 3 by 9 inches; all 
the mortices in the sills being 3 inches from the work sides. The 
general rule for draw bores and draw pins may be stated as follows:— 
The size of the draw bore should be equal to half the thickness of the 
tenon, when the tenon is not more than 3 inches thick; but it never 
need be more than 1J inches in size, even though the tenon may be 
more than 3 inches thick. In wide mortices, it is customary to have 
the tenons secured with two, and sometimes three pins, as represented 
in the Plate. Let one draw bore be 2 inches from one side of the 
mortice, and the other 2 inches from the other side, and each one 2 
inches from the face of the mortice. 

In the tenons, let the draw bores be 2 inches from each side, and 
about one fourth of an inch, in large tenons, nearer the shoulder than 
the draw bores of the mortices. Great care should be observed to 
have the draw bores perfectly plumb; and workmen should be cau¬ 
tioned against making a push bore , as it is called, when not plumb. 

(55) 


66 


CARPENTRY MADE EASY. 


The posts need not be pinned at the bottom, and the manner of pin¬ 
ning the other tenons is represented in the Plate. 

Braces. 

The braces are framed on a regular 3 feet run; that is, the brace 
mortice in the girder is 3 feet from the shoulder of the girder, and the 
brace mortice in the post is 3 feet below the girder mortice. Al- 
ways remember that the measure for braces and brace mortices is com¬ 
puted to the furthest end, or toe of the brace ) and the furthest end of 
the mortice. The mortices for 4 inch braces need to be 5} inches 
long, so that the end of the mortice in the post, next the girder, will 
be 2 feet 6J inches from the girder, and the end furthest from it will 
be 3 feet. The bevel of braces on a regular run is always at an angle 
of 45°, and is the same at both ends of the brace. 

Pitch of the Roof. 

In this building the roof is designed to have a third pitch; that is, 
the peak of the roof would be one third the width of the building 
higher than the top of the plates, provided the rafters were closely 
fitted to the plates at their outer surfaces, as in Plates Nos. 3, 4, and 6; 
but it is common in barns, and sometimes in other buildings, as has 
been already illustrated in Plate 5, Fig. 3, to let the rafters down only 
half their width upon the plates, allowing them to project beyond the 
plate, so that in this case the peak of the roof is 10 feet 3 inches above 
the plates, the pitch being still a third pitch, or 8 inches rise to a foot 
run. In order to give strength to the mortices for the upper end 
girders, these girders are framed into the corner post several inches 
below the shoulders of the post, say 4 inches; the thickness of the 
plates being 8 inches, it will be perceived that the dotted line, AB, 
drawn from the outer and upper corner of one plate to the outer and 
upper corner of the other, is just 1 foot higher than the upper surface 
of the girder; and that the peak of the roof is 11 feet 3 inches above 
this girder. The length and bevels of the rafters can be found as al¬ 
ready described in Plate 3 and Table 1. 

Purlins. 

The purlin plates should always be placed under the middle of the 
rafters; and the purlin posts, being always framed square with the 
purlin plates, the bevel at the foot of these posts will always be the 


BARN FRAMING. 


57 


same as tne upper end bevel of tbe rafters ;* also, the bevel at each 
end of the gable-end girder will be the same, since—the two girders 
being parallel, and the purlin post intersecting them—the alternate 
angles are equal. (Prop. XII., Cor., p. 29.) The length of the gable- 
end girder will be equal to half the width of the building, less 18 
inches; 6 inches being allowed for half the thickness of the purlin 
posts, and 6 inches more at each end for bringing it down below the 
shoulders of the posts. 


Length of the Purlin Posts. 

In order to obtain the length of the purlin posts, let the learner pay 
particular attention to the following explanation of Fig. 2. Let the 
point P represent the middle point of the rafter, and let the dotted 
line PO be drawn square with AB; then will AC be the J of AB, or 
7| feet, and PC, half the rise of the roof, will be 5 feet, and PO 6 
feet. The purlin post being square with the rafter, and PO being 
square with AB, we can assume that PR would be the rafter of an¬ 
other roof of the same pitch as this one, provided PO were half its 
width, and OR its rise; and then, since we know the length of PO, 
the length of PR could also be found by the rafter table (No. 1, Part 
IV.), as follows:—Width of building, 12 feet; rise of rafter, J of 12, or 
4 feet; hence, length of rafter, or PR, equals 7 feet 2 r 5 0 % inches; from 
this deduct half the width of the rafter and the thickness of the pur¬ 
lin plate, or 9 inches, and we have, 6 feet 5 T 5 5 2 5 inches as the length of 
the purlin post, from the shoulder at the top to the middle of the 
shoulder at the foot.f This demonstration determines also the place 
of the purlin post mortice in the girder; for AC being 7J feet, and 
OR being 4 feet, by adding these together, we find the point R, the 


* This fact is capable of a geometrical demonstration; for the triangle POR is similar 
to the triangle ACP; the side PR in one, being perpendicular to the side AP in the 
Dther, the side PO being also perpendicular to AC, and the side RO perpendicular to 
PC. (Part I., Prop. XXIX.) Hence, the angles opposite the perpendicular sides are 
equal; and we have angle APC, whieh is the same as the upper end bevel of the rafter 
—being parallel with it—eqval to PRC, the angle formed by the purlin post and the 
girder at their intersection at R. 

•}• The following geometrical demonstration of the above proposition is subjoined. In 
the two similar triangles ACP and POR, the sides about the equal angles are propor¬ 
tional (Def. 31); and we have, CP: AC: : OR: OP; but CP is § of AC; consequently, 
OR is | of OP. But OP equals 6 feet; hence, OR equals 4 feet. Again, the triangle 
POR being right-angled at 0, then PO a -f-OR 2 =PR 2 . 4 2 =16, and 6' J =36; 86-4-16=52, 
and y/62 P =7 ft 2.52 in., as above. 



58 


CARPENTRY MADE EASY. 


middle of the mortice, to be 11J feet from the outside of the build¬ 
ing; and the length of the mortice being 7J inches, the distance of the 
end of the mortice, next the centre of the building, is 11 feet 9f inches 
from the outside of the building. 

Purlin Post Brace. 

The brace of the purlin post must next be framed, and also the 
mortices for it, one in the purlin post and the other in the girder. 
The length of the brace and the lower end bevel of it will be the same 
as in a regular 8 feet run; and the upper end bevel would also be 
the same, provided the purlin post were to stand perpendicular to the 
girder; but, being square with the rafter, it departs further and fur¬ 
ther from a perpendicular, as the rafter approaches nearer and nearer 
toward a perpendicular; and the upper end bevel of the brace varies 
accordingly, approaching nearer and nearer to a right angle as the 
bevel at the foot of the post, or, what is the same thing, the upper end 
bevel of the rafter departs further and further from a right angle, 
lienee, the bevel at the top of this brace is a compound bevel, found by 
adding the lower end bevel of the brace to the upper end bevel of the rafter * 
(See Plate 8.) 


Purlin Post Brace Mortices. 

In framing the mortices for the purlin post braces, it is to be ob¬ 
served, also, that if the purlin post were perpendicular to the girder, 
the mortices would each of them be 8 feet from the heel of the post; 
but as the post always stands back, so the distance will always be 
more than 3 feet from the heel of the post; and the sharper the pitch 
of the roof, the greater this distance will be. Hence the true distance 
on the girder for the purlin post brace mortice is found by adding to 
3 feet the rise of the roof in running 3 feet; which, in this pitch of 8 
inches to the foot, is 2 feet more, making 5 feet, the true distance 
of the furthest end of the mortice from the heel of the purlin post. 

The place in the purlin post for the mortice for the upper end of 
the brace may be found from the rafter table, by assuming that ILc 


• This proposition is capable of demonstration, thus: The angle PxM equals the sum 
of the angles MRx and xMR, since PxM is the exterior angle of the triangle MRx, 
formed by producing the base Rx in the direction RxP. (See Prop. XVI., Cor.) But 
the angle PxM is the upper end bevel of the purlin post brace; therefore, it is equal to 
the sum of the two bevels, one at the foot of the brace and the other at the foot of the 
post, as above 





BARN' FRAMING. 


59 


would be the rafter of another' roof of the same pitch as this one, if 
xy were half the width, and ?/R the rise. For then, since xy equals 
3 feet, we should have width of building equal 6 feet, rise of rafter, 
one third pitch, gives ?/R equal 2 feet; and hence aR would equal 
3 feet 7.26 inches, the true distance of the upper end of the mortice 
from the heel of the purlin post * 


* The same proposition is demonsti*ated by Geometry, as follows: xy being parallel 
with PO, the two triangles 11PO and Rxy are similar, (Geom., Prop. XXIX), hence the sides 
opposite the equal angles are proportional, and we have Rx ; RP :: xy : PO. But we 
have already found PO to equal 6 feet, and xy equal to 3 feet, and RP equal to 7 feet 
‘V62 inches. Hence, 

6 s 8 :: 7 ft. 2.52 in. : 3 ft. 7.26 in. Answer as abovs. 



PLATE 8. 


UPPER END BEVEL OF PURLIN POST BRACES. 

Plate 8 is designed to illustrate tlie manner of finding the upper 
end bevel of purlin post braces, to which reference is made from the 
preceding Plate. 

In Fig. 1, let AB represent the extreme length of the brace from 
toe to toe, the bevel at the foot having been already cut at the proper 
angle of 45 degrees. Draw BO at the top of the brace, at the same 
bevel; then set a bevel square to the bevel of the upper end of the 
rafter, and add that bevel to BC, by placing the handle of the square 
upon BC and drawing BD on the tongue. This is the bevel required. 

Fig. 2 shows another method of obtaining the same bevel. Let the 
line AB represent the bevel at the foot of the brace, drawn at an angle 
of 45 degrees. Draw BD at right angles with AB, and draw BC per¬ 
pendicular to AD, making two right-angled triangles. Then divide 
the base of the inner one of these triangles into 12 equal parts, for 
the rise of the roof. Then place the bevel square upon the bevel AB, 
at B, and set it to the figure on the line CD, which corresponds with 
the pitch of the roof. This will set the square to the bevel required 
for the top of the brace. In this figure the bevel is not marked upon 
the brace, but the square is properly set for a pitch of 8 inches to the 
foot, or a one third pitch. The square can now be placed upon the 
top of the brace, and the bevel marked- 
( 60 ) 


Plate 8. 



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PLATES 9 & 10. 

Plates 9 and 10 exhibit the side and end elevations of a building 
designed for a warehouse, or mill. 

Length of building, 50 feet; 

Width of building, 40 feet; 

Height of building from the foundation to the top of plates-, 36 feet; 

Main timbers, 12 inches square; 

Door posts, 10 by 12 ; 

Purlin posts, 8 by 10; 

Plates and purlin plates, 8 by 8; 

Braces, 4 by 6 ; 

Lower joists, 8 by 12 ; 

Upper joists, 3 by 10; 

Studding, 2 by 8; 

Eafters, 2 by 6. 

The posts are framed in sections, one story at a time, on account 
of the difficulty in procuring long timbers, also for facility in raising 
the building; for, by this means, each story can be raised separately. 
It has also been proved by experience, that when the timbers are 
locked together as represented in the Plate, this mode of building is 
equally strong as to have the posts in one length. The ends of the 
joists are sized to a uniform width, and placed upon the timber 
ilie crowning side up ; the studs are morticed into the timbers P r ' 

The roof is framed to a quarter pitch, and the brace? ; a aea ar 
3 feet run. Plate 3 describes the manner of obtain-' a v eis of 
the rafters and gable-end studding. Plates 7 ^ ■.d 8 - i,ne manner 
of obtaining the bevels of the purlin po c,- ‘ or- •• o. Plate 4 gives 
the method of finding the length of the grle-eud studding. 

Cripple -.cuds. 

The length of the cripple studs , which are to be nailed to the braces, 
depends upon the run of the braces. The braces in this building, being 
on a regular run, are all set at an angle of 45 degrees, so the bevel of 
the cripple studs will be the same; and the rise of the brace being equal 
to the run, the length of each cripple stud will be equal to the height 
of the post from the sill to the toe of the brace, added to the distance 

(61) 


62 


CARPENTRY MADE EASY. 


of the stud from the post. In this building, the height of the brace 
from the sill to the toe of the brace in the first story is 8 feet, and the 
first stud being 14 inches from the post, and 2 inches thick, the length 
of the first cripple stud will be 16 inches longer than the height of 
the post from the sill to the toe of the brace, or 9 feet 4 inches; and 
the length of the next cripple stud will be 16 inches more, or 10 feet 
8 inches. 

It now remains to determine the bevels and the lengths of those 
cripple studs in the gable end, which are to come against the purlin 
posts. Having already (Plate 7) found the bevel at the foot of the 
purlin post equal to the upper end bevel of the rafters, it will follow 
that the bevel of the cripple studs upon the purlin post is equal to the 
lower end bevel of the rafters* The length of the cripple studs stand¬ 
ing between the rafter and the purlin posts depends both upon the rise 
of the roof and the rise of the purlin post; but the purlin post 
being set square with the rafter, its rise is always the same as the run 
of the rafter, and its run is the same as the rise of the rafter. 

Hence, for finding the length of a cripple stud, standing in any 
building between the rafter and the purlin post, at a certain horizontal 
distance from the top of the purlin plate, we have the following Rule: 
Add the RISE of the roof in RUNNING the given distance to the RUN of the rooj 
in RISING the given distance ; the sum will give the length of the cripple stud. 

For example, in this plate, suppose the cripple stud / to be 18 inches 
from the top of the purlin plate, horizontal distance, then the risb of 
the roof on a quarter pitch in running 18 inches would be 9 inches, 
?nd the run of the roof in rising 18 inches would be 36 inches; so that 
the 1 ength of I is 45 inches. The stud marked M being 16 inches 
from f the additional Vise is 8 inches, and the additional run is 32 
inch* s, so i.\ * M is 40 inches longer than I. 

Note on J 8eve ;' in a frame of this kind are only four in number:— 

1. The bevel o. the wi ■ ei r the rafter. 

2. The bevel of the » o> a ; 

3. The bevel of the braces, e equal to 45 degrees. 

4. The bevel of the upper end vf the purlin post brace, always equal to the sum of 
the first and third. Balloon frames have but two bevels—the first and second above 
mentioned. 


• Demonstrated as follows. The triangle ABC is similar to the triangle DEF, since 
the sides of the one are perpendicular to the sides of the other; consequently the angles 
opposite the perpendicular sides are equal. (Geom., Prop. 29.) 

The side FE, in one triangle, is perpendicular to the side BC in the other; hence, the 
angle A= angle D. The angle A is the lower bevel of the rafters, and the angle D is the 
bevel of the cripple stud on the purlin post. 

















































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PLATE 11. 


Plate 11 is designed to represent two modes of framing braces in a 
self-supporting or trussed partition. Where the span is considerable, there 
being no support beneath except the exterior wall, some mode of 
bracing is indispensable. These plans are exhibited as being practi¬ 
cable and secure. 

The first plan gives opportunities for two or even three openings. 

The second plan will be most convenient where only one opening 
is desired. 

The size of these brace timbers should be in proportion to the width 
of the building, and the weight which the partition is to sustain. If 
they are ten or twelve inches square, they will safely sustain a brick 
wall built upon the partition. 

Fig. 3 is designed to show the proper mode of trussing a beam over 
a barn floor, or in front of a church gallery, or any other situation 
where it is inconvenient to support it by posts. 


( 63 ) 


PLATE 12. 


SCARFING. 

This Plate exhibits several designs for scarfing or splicing timber. 
The length of the splice should be about four times the thickness of 
the timber; and when the joint is beveling, it will be found the best 
and most expeditious way, first, to prepare an exact pattern of boards, 
and then to frame the timbers by the pattern: by this means a perfect 
ioint can be made. 


Straps and Bolts. 

Pig. 4 is spliced by strapping pieces of plank upon the upper and 
lower sides of the joint, and securing them with bolts of f inch or 1 
inch in diameter, according to the size of the timber. 

Pigs. 5, 7, and 8, have iron straps bolted in a similar manner. 

Fig. 9 exhibits a strong mode of splicing timbers where they are 
doubled throughout their whole length, for a very long span, such as 
roofs in churches. 

Those styles which are numbered 1, 2, 7, and 5, are recommended 
as being the best in proportion to the cost. 

(64) 






Plate 12. 



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Plate 13. 










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PLATE 13. 


Scarfing, whenever it is practicable, should be made directly over a 
post, when a simple and inexpensive style, such as is exhibited in Plate 
13, will be found sufficiently strong. Indeed, it is hardly possible 
to find a stronger mode of scarfing than that illustrated in Fig. 1; and 
yet, being supported by the post, the design is more simple than most 
of those represented in Plate 12. 

In this design, the head of the post is framed into a bolster, which 
ought to be fully equal in size to the timber which it is required to 
support; in heavy frames it should be about 6 feet long, the braces 
being framed on a 4 feet run. The bolster is secured to the timber 
with inch bolts, as represented in the figure. 

Fig. 1 also illustrates the proper mode of framing braces; the dotted 
lines show the form of the tenons, and the notchings in the post and 
the girder represent the facing of the mortices. These are usually 
notched-in half an inch at least, in order to give all possible support 
to the toe of the brace; and the measurement, for the length and the 
run of the braces, must be from the furthest point of the face of the 
mortice, inside of the notch, and not from the outside of the rough 
timber. 

The shoulder of the post is also notched or sized into the bolster an 
inch, and the bolster is locked into the girder 2 inches; and, if the 
timber is 12 inches square, the shoulder of the post is 10 inches lower 
than it would be without any bolster; and the carpenter must, of 
course, make his brace mortices in the post, accordingly, 10 inches 
nearer the shoulder than usual. 

Figs. 2 and 3 are designed to represent a less expensive mode of 
scarfing on posts. In these plans the tenons extend quite through, 
and are double-pinned to both timbers, as represented in the plate. 
This mode is sufficiently strong for scarfing plates and purlin plates. 

5 ' (65) 


/ 


PLATE 14. 


FLOORS IN A BRICK BUILDING. 

Plate 14, Fig. 1, exhibits the ground plan of one room, 18 feet wila, 
'file joists are 2 by 10, 18 feet long. 

Trimmer Joists. 

Those marked A, B, and C, at each side of the fire-place, are called 
trimmer joists; they are 4 by 10; or they may be made by spiking two 
common joists together, as represented in the plate. 

A course of bridging is represented at D, as described in Plate 6, 

Fig. 2 and Fig. 5 exhibit the method of framing the tenons of the 
common joists. 

Fig. 3 shows the mode of framing together the trimmer joists at the 
corners of the hearth 

Fig. 4 shows the beveled ends of the joists where they are set into 
the brick wall. They are beveled in the manner represented, that 
the springing of the joists may not endanger the wall; and, in case of 
fire, the joists may burn and fall out without destroying the wall. 

( 66 ) 























































































































































































































































































PLATE 15. 


CIRCULAR CENTRES. 

Plate 15 exhibits several designs, more or less convex, for the con¬ 
struction of centres, which are skeletons used by stone and brick 
masons to build an arch upon, but which are to be taken away when 
the arch is sprung and the mortar set. 

First, draw the line AB upon a floor, two inches less than the width 
of the skeleton required, and take OA as a radius, and describe the 
semi-circumference. Inch boards are then to be fitted to this curved 
line, and their ends beveled to fit each other, as represented in the 
Plate. The bevel is determined by simply drawing a straight line from 
any point of the curve to the central point 0. Other boards are then 
to be nailed over the joints, on the inside of these ; and the long brace 
AB nailed at each end at the bottom. 

Having prepared the other end of the skeleton in the same manner, 
strips of boards, an inch thick, two inches wide, and of a length equal 
to the thickness of the wall, are nailed upon their convex edges, as 
represented. 

Should the arch have more than 12 feet span, it would be proper 
to use thicker boards; but for any thing less than 12 feet, inch boards 
are amply sufficient. 






PLATE 16. 


ELLIPTICAL CENTRES. 

This Plate illustrates the manner of constructing elliptical centres. 
The elliptical curve is described most accurately by means of a tram- 
mel, the construction and use of which are explained in Plate 2, page 20. 

In order to describe the curve for these centres, take AB equal to 
the span of the arch, less 2 inches, and set the trammel so that the 
intersection of the arms will fall upon 0, the middle point of AB. 
Then set the pin B, so that PB will equal the height of the. arch less 
1 inch; and set the pin C so that PC will equal AO. 

( 68 ) 




































































































































































































PLATE 17. 

ARCHES. 

It is also the business of the carpenter to prepare patterns for stone¬ 
cutters, by which they are to cut their stones to fit arches of any de¬ 
sired form. This Plate exhibits seven different styles of arches, with 
the most accurate and convenient modes of drawing them, and of 
dividing them into proper sections or patterns for the arch-stones. 

Fig. 1 represents an elliptical arch, drawn by means of a trammel, 
as has been already described. The arch is divided into blocks of 
proper form and size, by first dividing the curve into any desired 
number of equal spaces; then, wherever a joint is required, first draw 
a tangent to the curve at that point, and then a line perpendicular to 
this tangent will divide the arch properly. 

Fig. 2 exhibits the Tudor arch, drawn in two rampant semi-elliptical 
curves, as represented in the Plate. The same rule is to be observed 
as before for finding the joints of this arch, and for dividing it cor¬ 
rectly into proper patterns. 

Fig. 3 is a semicircular arch. This is most easily and correctly 
jointed by drawing a radius from the centre 0 to any point of the 
curve where a joint is desired. 

Fig. 4 is a Gothic arch, described by two equal radii from the 
points 0, O, as centres, and jointed from the same points. 

Fig. 5 is so similar to the last, as to require no further description. 

Figs. 6 and 7 exhibit two depressed segmental arches of the same 
span, but representing different degrees of curvature, one being drawn 
with a longer radius than the other, and both jointed by radii drawn 
from the common centre 0. 


( 69 ) 


PLATE 18. 


HIP ROOFS. 

Plate 18, Fig. 1, exhibits the plan of a hip roof in a building 50 feet 
long and 40 feet wide, the rise of the roof being 5 inches to the foot. 
AB, CD, and EF, are the upper girders, which are trussed for support¬ 
ing the roof, with short principal rafters and straining beams, as rep¬ 
resented in the Plate at CD. 

GH, III, KL, and ML, are the hip rafters, and IIL the ridge pole. 
The purlin plates are placed upon the principal rafters, and the strain¬ 
ing beams are framed on a level with the purlin plates, so that the 
end ones may answer the double purpose of straining beams and 
purlin plates also. 

The lengths* and bevels of the common rafters, in the middle of the 
building, the upper ends of which rest against the ridge pole, are 
found as usual in common roofs. (Table No. 1.) 

Hip Rafters. 

The length of the hip rafters is given in the Ilip Rafter Table, (No. 
2, p. 133) where the rule for obtaining it is fully explained. 

In addition to the two bevels common to all rafters, namely, the 
upper end bevel and the lower end bevel, hip rafters have two other 
bevels, which are the side bevel , as it is called, being the angle which 
the two hip rafters make with each other at their intersection, and the 
backing, which is the angle made by the intersection of the side roof 
with the end roof. 

Side Bevel of the Hip Rafters. 

It is evident that if there were no pitch to the roof, this bevel would 
be 45 degrees, since the two hip rafters would be perfectly square with 
each other; but as soon as the roof begins to rise, the hip rafters are 
no longer square with each other, but begin to approach a parallel. 


• Throughout this work, the length of rafters is computed from the upper and outer 
corners of the plates to the very peak of the roof, without allowances for the projection 
of the rafters or the thickness of the ridge poles. In case the building has a ridge pole, 
therefore, it will be necessary to deduct one half the thickness of it from the length as 
given, measuring for this deduction square from the down bevel, and not lengthwise of 
the rafter. 

( 70 ) 




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HIP HOOFS. 


71 


till, in a very steep roof, as that of a tower or steeple, they are nearer 
parallel than square with each other. The side bevel, therefore, is 
always greater than 45 degrees, and is obtained from the square by 
taking the length of the hip rafter on the blade, and its run on the 
tongue; the blade then shows the side bevel of the hip rafter re¬ 
quired. 

Down Bevel of the Hip Rafters. 

The upper end bevel is commonly called, in hip rafters, the down 
bevel. It is always square with the lower end bevel, the one being 
the complement of the other. This bevel varies with the pitch of the 
roof; for the pitch of the hip rafter always has the same propor¬ 
tion to the pitch of the common rafter, that the run of the common 
raftei has to the run of the hip rafter, or that the side of a square has 
to its diagonal; for, if we let 0 represent the foot of the perpendicular 
let fall from L upon the middle of the girder, CD, then ODPM is a 
square, of which OD, the run of a common rafter, is the side, and OM, 
the run of the hip rafter, is the diagonal. Now, it is a well-known 
principle in mathematics, that the side of any square is proportioned to 
its diagonal , as ONE is to the SQUARE ROOT OF TWO, (Prop. XXIV., 
Cor.); or, as 1 is to 1.4142; or, as 12 inches are to 17 inches nearly. 
AVhen the common rafter, therefore, has 5 inches rise to 12 inches run, 
the hip rafter of the same roof has 5 inches rise to 17 inches run; and 
when the common rafter has 6 inches rise to 12 inches run, the hip 
rafter has 6 inches rise to 17 inches run; &c. 

From these demonstrations we derive the following Rule for find¬ 
ing the down bevel and the lower end bevel of the hip rafters. Take 
17 inclces on the blade of the square for the run, and the rise of the roof to 
the fool on the tongue. The tongue will give the down bevel, or upper 
end bevel, and the blade the lower end bevel. 

Backing of the Hip Rafters. 

This is found on the square by taking the length of the hip rafter 
on the blade, and the rise of the roof on the tongue; the bevel of the 
tongue will be the backing required. For illustration, in this build¬ 
ing the length of the hip rafters, as given in the Table, equals 29 J 
feet nearly , and the rise of the roof is 8} feet. Take proportional parts 
of each on the square, and the tongue will give the backing; that is, 
place the square upon a strait edge, with the blade at the 14f inch 
mark, and the tongue at tne 4J- inch mark, and draw a scratch along 
the tongue; then set a bevel square to the angle which this scratch 
makes with the straight edge, and it is the backing required. 

For further illustration see Plate 39, and page 111. 


72 


CARPENTRY MADE EASY. 


Lengths and Bevels of the Jack Rafters. 

Since the pitch of the jack rafters is the same' as that of the common 
rafters, the longest jack rafter, the upper end of which rests against 
the end of the ridge pole, is of the same length as the common rafters, 
as given in the Common Rafter Table, less half the thickness of the 
hip rafters at their side bevel on the ridge pole. The difference in 
length between the longest jack rafter and the next one, or between 
any two adjacent ones, is equal to their distance apart, added to the 
gain of the rafter in running that distance. 

For example, in this building, the width being 40 feet, the run of 
the jack rafter is 20 feet, or 240 inches; and its length , on a pitch of 5 
inches rise to the foot, is 260 inches; therefore, its gain is 20 inches in 
running 20 feet, or an inch to a foot. The jack rafters being 2 feet 
apart, the difference in length between any two adjacent ones is, there¬ 
fore, 2 feet 2 inches. 

Or, the length of the shorter jack rafters may be obtained from that 
of the longest one, by dividing the length of the longest one by the 
number of spaces between the longest one and the corner of the build¬ 
ing ; the quotient will be the length of the shortest one, and it will be 
also the difference between any two adjacent ones. 

The down bevel and the lower end bevel are the same as the upper and 
lower end bevels of the common rafters. 

The side bevel of the jack rafters is always more than 45°, for a simi- 
lar reason as that given in the description of the side bevel of the hip 
rafters; and, since all the jack rafters have the same pitch as the com¬ 
mon rafters, we have the following 

Rule for obtaining the side bevel of the jack rafters .—Take the length of 
a common rafter on the blade of a square, and its run on the tongue— 
or proportional parts of each. The bevel on the blade is the side bevel 
of all the jack rafters in the frame. 

Or, take the length of the longest jack rafter on the blade, and half 
the width of the building on the tongue, or proportional parts of each, 
and the bevel on the blade will be the required side bevel. 

Remark. —In a hip roof which is perfectly square, the hip rafters 
need have no side bevel; for two of them can be cut of full length, 
and set up, opposite each other first, with their down bevels resting 
full against each other like common rafters; the other two hip rafters 
can then be set against these, having been cut off* half their thickness 
shorter than the full length, 

































































































































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PLATE 19. 


HIPS AND VALLEYS. 

Plate 19, Fig. 1, represents the hip roof of a building consisting of 
a main portion and a wing, the walls of the wing being of the same 
height as those of the main building. The main building being 24 
by 30 feet, and the wing 10 by 20 feet, the roof of the wing being of 
the same pitch, will not rise as high as that of the other part. 

The timbers AB and CD at the intersection of these roofs are called 
valley rafters. The upper end of the first one, AB, is extended to the 
ridge pole of the main building; the other valley rafter is supported 
by the first one, by being spiked to it at their intersection at C. From 
that point of intersection, a ridge pole extends to the intersection of 
the end hip rafters at E. This ridge pole is equal in length to the 
whole width of the wing; the valley rafter and hip rafter on each side 
being parallel with each other. 

The lengths and bevels of the various rafters are found as explained 
in the preceding Plate. 

TRAPEZOIDAL HIP ROOFS. 

Fig. 2 exhibits the plan of a hip roof for a building constructed in 
the form of an irregular square, or trapezoid ; the side CD being 4 feet 
longer than the side AB, the width 24 feet and the rise 5 inches to the 
foot. 

The length and bevels of the common rafters and of the hip rafters 
on the square end of the frame are obtained in the same manner as de¬ 
scribed, in the regular hip roof. The lengths and bevels of the two 
hip rafters BF and DF on the beveled end of the frame are unlike each 
other, and unlike those on the square end, one being longer and the 
other shorter than those. As such buildings are comparatively rare, it 
has not been deemed necessary to encumber this work with a table 
« for such rafters, but the facts and principles applicable to such frames 
are here stated. 


Lengths of the Irregular Hip Rafters. 

If this end of the building were square, then BG and DI would 
each of them be equal to half the width of the building, or 12 feet; 

(73) 


74 


CARPENTRY MADE EASY. 


but since one side of the building is 4 feet shorter than the other side, 
the short side is 2 feet less than 12, and the long side 2 feet more than 
12 ; or BG equals 10 feet, and DI equals 14 feet. 

The length of the common rafters GF and IF is found by the table 
to equal 13 feet; and the two triangles GFB and IFD being right- 
angled, we are now furnished with the means of finding the lengths of 
the two hip rafters BF and DF, by the use of that familiar principle in 
mathematics, that in every right-angled triangle the square of the hypote¬ 
nuse is equal to the sum of the squares of the other two sides. (Part I., Prop. 
XXIY.) For in the right-angled triangle GFB, we have the side GB 
equal to 10 feet, which we reduce to inches, to secure greater accuracy in 
calculation. GB then equals 120 inches, and GF, a common rafter, 
equals 13 feet or 156 inches. The square of 120 is 14,400, and the 
square of 156 is 24,336; the sum of these squares is 38,736, which is 
the same as the square of the hypotenuse BF; and by extracting the 
square root of 38,736 inches, we have 195 inches and 81 hundredths of 
an inch, or 16 feet 4.81 inches, as the length of the hip rafter BF. 

The length of BF is found in a similar manner from the triahgle IFB, 
DI being 14 feet or 168 inches, the square of which is 28,224, to which 
add the square of IF, equal to that of GF, found above to be 24,336. 
The sum of these squares is 52,560 inches, of which the square root is 
229.25 inches, or 19 feet 1.25 inches, the true length of the longest hip 
rafter. 


Bevels of the Irregular Hip Rafters. 

The down bevel and the lower end bevel of these irregular hip 
rafters, like those of all other rafters, are square with each other: and 
are found together on the square by taking the run on the blade and 
the rise on the tongue ; the tongue will give the down bevel, and the 
blade the lower end bevel. 

Note .—Take the square of BG, and the square of DI, respectively; add to each the 
square of half the width of the building; extract the square root of each of these sums, 
and they will give the runs of BF and DF, respectively. 

Thus, the square of BG, 10 feet, or 120 inches, is 14,400 inches; the square of DT, 14 
faet, or 168 inches, is 28,224 inches; and the square of half the width of the building, 
12 feet, or 144 inches, is 20,736 inches; this added to 14,400 is 35,136 inches, the square 
root of which is 187.44 inches, or 15 feet 7.44 inches, which is the run of BF. 

Again, 20,736 added to 28,224 equals 48,960, the square root of which is 221.26 inches, 
or 18 feet 5.26 inches, the run of DF. 

The side bevel of irregular hip rafters is obtained by adding together the 
distance from the foot of the hip rafter to the foot of the first common 


TRAPEZOIDAL HIP ROOFS. 


75 


rafter, and the gain of the hip rafter; then take this sum on the tongue 
of a square, and half the width of the building on the blade; and the 
tongue will give the side bevel required. 

Backing of Hip Rafters on Trapezoidal and other Irregular Roofs. 

As the oblique angles of trapezoidal and other irregular buildings 
are liable to many variations, so the backing of the hip rafters must 
also vary on different roofs. The simplest and most practical manner 
of finding these backings, when thus irregular, is to take a small square 
block, and bevel one end of it to the same bevel as the lower end of 
the hip rafter in question. Then place this beveled end upon the 
plates, just as the hip rafter is to be placed, at the oblique corner of 
the building, and draw a pencil line on the under side of the block, 
along the upper and outer edges of the plates, till they meet at the 
corner; these lines will be the bevel of the backing required. And 
then the block can be worked off to the lines, and a bevel square set 
to the angle thus formed, by which the hip rafter itself can then be 
beveled. 


Length of the Jack Rafters. 

First. Uf those on the long side of the building, between D and I. 
In the triangle DFI, the jack rafter f being parallel with the base FI, 
has the same proportion to FI that D f has to D1 (Geom., Props. XXY. 
& XXVII.) ; and in order to find the length of the jack rafters coming 
between I and D, we have only to divide IF by the number of rafters 
required from IF to the corner inclusively, and the quotient will be 
the difference between any two of them, and will equal also the length 
of the shortest one. Since it is 14 feet from I to the corner of the 
building, and the rafters are 2 feet apart, it will require 7 rafters, in¬ 
cluding IF. We therefore divide IF, 13 feet, by 7, and have 1 foot 
lOf inches for the difference between IF and f and also between /and 
e; or, what is equally true, we can take this number, 1 foot 10| inches, 
as the length of a; double this, or 3 feet 8$ inches, for the length of 
b ; three times this number, or 5 feet 6? inches, for the length of c; 7 
feet 5^ inches, the length of d; 9 feet 3f inches, the length of e; 11 
feet If inches, the length of /; and 13 feet the length of IF: proving 
the calculation to be correct. It will be most convenient in practice 
to cut the longest ones first, that the short pieces of stuff may be 
worked in to better advantage. 

Second. In the same manner obtain the difference between GF and 


76 


CARPENTRY MADE EASY. 


on the other side of the building, by dividing GF, 13 feet, by 5, tne 
number of rafters required in GB; 13 feet divided by 5 equals 2 feet 
7 1 inches; this is at once the length of g and the difference between 
any two adjacent rafters between B and G. 

Third. Of those on the end of the building. Divide the length 
of HF by 6, which is the number of rafters required between I1F 
and each corner of the building, including HF. IIF equals the 
length of a common rafter, less 2 inches—the bevel of the hip rafters 
•—or 12 feet 10 inches, which, divided by 6, equals 2 feet If inches, 
the length of each of the short rafters on the end, and also the differ¬ 
ence between HF and the jack rafters on each side of I1F. 

Side Bevels of the Jack Rafters on the Sides of the Frame. 

It is evident that, if the roof were horizontal, this bevel would be 
found on the square, by taking half the width of the building on the 
blade, and the distance from the corner of the building to the foot of 
the first common rafter on the tongue; but when the roof begins to 
pitch, this bevel will be too short; and the relation of the pitch of the 
roof to the length of the common rafter is such, that the side bevel of 
the jack rafters is obtained with perfect accuracy, by taking the length 
of the common rafter on the blade, and the distance from D to I and from 
B to G , respectively , on the tongue —the blade will give the side bevels 
required. 

The down bevels and lower end bevels of these jack rafters on the 
sides of the frame are the same as those of the common rafters, since 
they have the same pitch that the common rafters have. 

Side Bevels of the Jack Rafters on the Slant End of the Frame. 

For a similar reason assigned above, add to BG the gain of a common 
rafter in running the length of GB. Then take this sum on the blade of a 
square , and half the width of the building on the tongue, and it will give 
the side bevel of the end rafters which are nearest BG. In like man¬ 
ner, add to DI the gain of a common rafter in running the length of 
DI. Take this sum on the blade, and half the width of the building 
on the tongue, and it will give the bevel required of those rafters 
nearest DI. 

For illustration, BG equals 10 feet; as the common rafters are 1? 
feet long, in running 12 feet, they gain an inch in running a foot. So, 
m running 10 feet, a rafter would gain 10 inches. We therefore take 
u certain proportional part of 10 feet 10 inches on the blade of a 


TRAPEZOIDAL HIP ROOFS. 


77 


square, say lOf inches, and the same proportional of half the width of 
the building, say 12 inches, on the tongue; the bevel of the blade is 
the side bevel for all those jack rafters on the end of the frame which 
are nearest GB, or which rest against BF. 

So, also, DI equals 14 feet, to which add 14 inches; hence, take 
15g inches on the blade, and 12 inches on the tongue; the bevel on 
the blade will be the side bevels of those jack rafters on the end of 
the frame nearest DT, which rest against the hip rafter DF. 

Down Bevel of the Jack Rafters on the Beveled End of the Frame. 

In order to obtain these bevels and the corresponding lower end 
bevels with perfect accuracy, it is necessary to obtain them for each 
rafter separately, for there are no two of them which have the same 
pitch. In order to ascertain with ease what this pitch and the corres¬ 
ponding bevels are, it is necessary to suppose all these end rafters to 
be produced beyond where they now rest on the hip rafters, and all 
to rest upon a common ridge pole PR, which is drawn through the 
point F, and extends on a level with F directly over the points G and 
I. These rafters would then have each one the same pitch they now 
have, since we have not supposed their direction changed, but their 
length only. Now, however, the rise is 5 feet for each one, and their 
run can easily be computed from the length of BG; for the run of the 
one nearest B is 4 inches more than BG, or 10 feet 4 inches; that of 
the next one, 4 inches more, or 10 feet 8 inches; &c. 

Now, in order to obtain the down bevel and the lower end bevel of 
any rafter, we take the run on the blade of a square, and the rise on 
the tongue. 

The measurements of these bevels are therefore as follows: 

5 inches on the tongue for each of them, and lOf inches, 10§ inches, 
11 inches, 11J- inches, Ilf inches, 12 inches, 12f inches, 12f inches, 
13 inches, 13 J inches, 13 f inches, respectively, on the blade. 


PLATE 20. 


OCTAGONAL AND HEXAGONAL ROOFS. 

Plate 20, Fig. 1, is designed to exhibit the proper mode of framing 
the roof of a building, the ground plan of which is a regular octagon. 
This style of building having become quite common, a Eafter Table 
(No. 3) has been prepared, which will be found very useful and con¬ 
venient, as it gives at one view the precise lengths of the longest hip 
rafters and the longest jack rafters. In the introduction or explana¬ 
tion of the table, full instruction and demonstrations are given for 
enabling the intelligent mechanic to test these calculations, or to ex¬ 
tend and apply them to roofs of other dimensions. 

Length of the Hip Rafters. 

The length given in the Table is that of the first pair, and is cal¬ 
culated from the outer and upper corner of the plates, at their inter¬ 
section with each other, to the very central point or apex of the 
roof. 

One of the first pair having been cut off of the above length and of 
the proper down bevel, the length of the second pair is obtained from 
this one, by taking off from this one half its thickness, measured back 
square from the down bevel; and the length of the third and fourth 
pairs, by taking off in like manner two thirds its thickness, or more 
accurately, \\ of its thickness, since 17 is half the diagonal of a square, 
the side of which is 24. (See Prop. XXIV., Cor) 

Bevels of the Hip Rafters. 

The doiun bevel , and lower end bevel, are found as usual, that is, by 
the run of the rafter and its rise, on the square. (The run of the lap 
rafter is half the diagonal width of the building. See Explanation of 
Octagonal Hip Roof Table.) 

The first and second pairs of hip rafters have no side bevel , since, 
in raising the frame, the first pair, HI, have their ends resting against 
each other, as in common rafters. The second pair, AP, are next 
raised at right angles with the first, and their ends resting square 
against the first. The third and fourth pairs have their side bevels 
(78) 
































































































































































































































' 














* 
















































































































































































































OCTAGONAL AND HEXAGONAL EOOFS. 


79 


cut on both sides of the upper end; and that bevel is found by taking 
the length of the rafter and its run on the square. 

The backing of the hip rafter is obtained on the square by taking j 5 2 
of its rise on the tongue, and its length on the blade. 

Length of the Jack Rafters. 

The length of the middle jack rafters is given in the Table. Having 
found this, in order to obtain the length of the shortest one, proceed as 
directed in the article on trapezoidal hip roofs; that is, divide the 
length of the longest one by the number of rafters required between 
that and the corner inclusively—the quotient will be the length of the 
shortest one, and also the difference between any two adjacent ones. 

The bevels of the jack rafters are obtained on the same prin¬ 
ciple as those in common hip roofs. 

The down bevel is found on the square by taking the length and the 
run of the middle jack rafter; and the side bevel by taking the length 
of this rafter and half the side of the building. 

Fig. 2 represents the roof of a building, the ground plan of which 
is a regular hexagon. 


Width of the Building. 

In every regular hexagon the side is equal to the radius of the cir¬ 
cumscribed circle; and the diagonal width is equal to the diameter of 
the circumscribed circle, or twice the side. If we let O' represent the 
foot of the perpendicular let fall from 0, then AB and BE are each 
equal to AO' and BO', and AD is equal to twice AB.* 

Half the square width of the building is found by subtracting from 
the square of the side, the square of half the side, and extracting the 
square root of the difference: since, in the right-angled triangle CO'B, 
CO' 2 =B0' 2 —CB 2 . The side of this building being supposed to be 20 
feet, we have the square of 20=400, and the square of 2 2°=100; their 
difference is 300 feet, the square root of which is 17.35 feet, or 17 feet 
4.2 inches, which is half the width of the building, or the run of the 
middle jack rafter 


The Length of the Rafter 

Must be computed as explained in the Introduction to the Rafter 


* See Geom., Prop. XXXI. 



80 


CARPENTRY MADE EASY. 


Table; that is, to say, every rafter is the hypotenuse of a right-angled 
triangle, of which its run and its rise are the other two sides. The 
square of the length is therefore equal to the sum of the squares of 
the run and the rise. 

In this Fig. the roof is supposed to rise 8 inches to the foot. The 
whole rise is therefore one fourth the run of the middle jack rafter, or 
one eighth the square width of the building, or 4 feet 4.05 inches, the 
square of which is 2,709.2025 inches. The run of the hip rafter we 
have already proved to be equal to the side of the building, or 20 
feet=240 inches, the square of which is 57,600 inches. The square 
root of the sum of the above two numbers is 245.56 inches, or 20 feet 
5.56 inches, which is the length of the hip rafter. 

In a similar manner the length of the middle jack rafters is found to 
be 17 feet 10.26 inches. The length of the shortest jack rafters is 
obtained in the same manner as in hip roofs generally. 

The bevels of the hip jack rafters are obtained by the same rule as in 
octagonal roofs. 

The backing of the hip rafters is found on the square by taking of 
its rise on the tongue, and its length on the blade. 





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V Inlo 21. 



Th t o.T.nin'Ka vd t A S w Phil a. • 















































































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PLATE 21. 

ROOFS OF BRICK AND STONE BUILDINGS. 

Plate 21, Fig. 1, exhibits a simple and excellent mode of framing 
a roof of a moderate span of from 25 to 35 feet, designed for a stone 
or brick building. The iron bolts, by which the feet of the princi¬ 
pal rafters are secured to the tie-beam, are f of an inch in di¬ 
ameter, and the supporting rod in the middle is 1 inch in diameter. 
The block between the upper ends of the principal rafters, is beveled 
to suit the pitch of the roof, while the ends of the principal rafters 
are square. The block should be of hard wood, and placed with its 
grain running in the same direction as that of the rafters, to avoid 
shrinkage. 

By means of the supporting rods, any proper degree of camber or 
crowning can be given to the tie-beam. 

The bolts and rods in this and the following figures are left rough 
in the Plates, to show them more distinctly ; but if the rooms beneath 
are to be ceiled or plastered, the heads of the bolts should be counter¬ 
sunk into the beam, and the nuts screwed upon the upper ends. 

Fig. 2 represents another style of roof, with two braces and two 
additional rods to each bent, and also a 2 inch block of hard wood 
about 2 feet long, placed between the feet of the braces. The grain 
of this block should also run parallel with the beam. 

Length and Bevels of the Braces. 

Suppose the distance of the upper end of the brace mortice in the 
principal rafter from the heel of that rafter to be five ninths of the 
whole length of that rafter, then the perpendicular let fall from the 
heel of the brace to the tie-beam will be five ninths of the rise of the 
rafter—and this we will call the rise of the brace ; then also the distance 
from the foot of this perpendicular to the toe of the foot of the brace, 
will be four ninths of one half the length of the tie-beam less 2 feet— 
one foot for the setting back of the principal rafter, and one foot for 
half the block at the foot of the brace: this is the run of the brace ; then 
the sum of the squares of the rise and the run will give the square of 
6 (81, 


82 


CARPENTRY MADE EASY. 


the length of the brace, the square root of which will be that length 
from the lower toe to the upper heel. 

The Span of this Roof 

May be from 50 to 65 feet. The middle rod should be 1J inches, 
the others f of an inch. 

Dimensions of Timbers for Figs. 1 & 2. 

For both frames —Rafters, 2 by 6; plates and purlin plates, 6 by 8. 

For each bent in Fig. 1—Tie-beams, 8 by 10; principal rafters, 7 by 9. 

For each bent in Fig. 2—Tie-beams, 10 by 12 ; principal rafters, 7 by 
10; braces, 6 by 7. 

The bents in each frame may be from 10 to 14 feet apart. 

Fig. 3 represents a simple mode of framing a roof for a shop or 
foundry where it is required to have a ventilator. 




























































PLATE 22. 


Pig. 1 represents a bent of a very strong roof, designed for machine 
shops and other buildings, where it is necessary for great weights or 
heavy machinery to be supported by them. The Author of this work 
has superintended the construction of roofs of this kind upon the ma¬ 
chine shops for the Railroad Works at Peoria, Ill., and at Davenport, 
Iowa, where they have proved sufficiently strong to sustain locomo¬ 
tive engines weighing more than twenty tons, when hoisted from the 
ground, and suspended by chains from the roof for repairs. 

The width of the building represented in Fig. 1 is 50 feet; the prin¬ 
cipal rafter is set back a foot from the end of the tie-beam, to give 
room for the wall plate; the rise of the roof is 5 inches to the foot. 
In framing roofs of this kind, the supporting rods should be furnished 
before commencing the frame: for then the length of the short prin¬ 
cipal rafters and that of the straining beam, can be regulated or pro 
portioned according to the length of the rods. It is best, however, 
for the middle rod to be twice the length of the short ones, reckoning 
from the upper surface of the beam to the upper surface of the prin¬ 
cipal rafters, and allowing 1 foot more to each rod for the thickness 
of the beam, and the nut and washer. For example, the middle rod 
is 11 feet long, and the short ones 6 feet each; which, after allowing 
1 foot, as above mentioned, makes the length of the long one, above 
the work side of the beam, twice that of the short ones. 

The length of the rod above the beam is the rise of the rafter, and 
the distance from the centre of the rod to the foot of the rafter is the 
run of the rafter; the length of the rafter can therefore be found from 
the Common Rafter Table. 

Length of the Straining Beam. 

Add the run of the short principal rafter to the lower end bevel of 
the long one ; subtract this sum from the run of the long principal, 
and the difference will be half the length of the straining beam. 

The bolsters under the ends of the tie beams are of the same thick¬ 
ness as that, and about 5 feet long. 

Fig. 2 is in every respect similar to Fig. 1, with the exception 
of the long principal rafters and the middle supporting rod being 
omitted. This roof is suitable for blacksmith shops and foundries, 
as it is also capable of sustaining great weights, and may be very con¬ 
venient for the safe storing of unwrought iron bars upon the beams. 

(83) 


PLATE 23. 


This plate exhibits sections of three roofs, of different dimensions, 
but similar to each other in style. This style is ancient, and, no doubt, 
has been proved of sufficient strength; but it is not recommended for 
convenience or economy, except where labor is cheap, timber plenti- 
ful, and iron scarce. 

In Fig. 1, the post in the middle is called the king post , and the 
other two queen posts. The tie beam is secured to their feet by iron 
straps; and braces extend in pairs from the posts to the principal 
rafters, as represented in the Plate. The heads of the posts are bev¬ 
eled to correspond with the pitch of the roof, and the ends of the prin¬ 
cipal rafters are left square. 

The manner of scarfing the tie beam is represented immediately 
below. 

Fig. 2 is very similar in design to Fig. 1, in the preceding Plate, 
the principal difference consisting in having king and queen posts 
instead of supporting rods. 

Fig. 3 shows the ridge pole supported by braces from the ends of 
the straining beam. The king post is, therefore, omitted; and the 
space between the queen posts may be appropriated for an attic cham¬ 
ber. The queen posts are let into the tie-beam an inch or more, to 
prevent displacement by the lateral pressure of the braces. 

(84) 































































































































































































































Plate *24. 


X 






N 


* X 


Uk-o.! cnithurdf KSon i’lula. 































































































































































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PLATE 24. 


Plate 24 exhibits several designs for rot>fs in a new and improved 
style, particularly adapted to those of a great span, as they may be 
safely extended to a very considerable width, with less increase of 
weight, and less proportionate expense, than any of the older styles. 
The principle on which they are constructed is essentially the same 
as that of the Howe Bridge. The braces are square at the ends, the 
hard wood blocks between them being beveled and placed as described 
in the foregoing Plates. Each truss of this frame supports a purlin 
post and plate, as represented. 

These roofs are easily made nearly flat, and thereby adapted to 
metallic covering, by carrying the walls above the tie beams to any 
desired height, without altering the pitch of the principal rafters, 
which ought to have a rise of at least 4 inches to the foot, to give a--' 
sufficient brace to the upper chord or straining beam. / 

f 

Fig. 1 is represented with counter-braces; and j 

\ 

Fig. 2 without them. The counter-braces do not add any thing'- to 
the mere support of the roof, and are entirely unnecessary in frames 
of churches, or other public buildings, where there is no jar; but they 
may very properly be used in mill frames, or other buildings designed 
for heavy machinery. 


(86) 


A 





PLATE 25. 




This Plate exhibits two plans for roofs of the same style as the last, 
but of simpler construction, and designed for a shorter span. 



In Fig. 2 the middle truss is omitted, to afford room for an attio 
chamber. 

( 86 ) 
























































































































■ 

. 












' 





























































( 





















m fl |iE k 111 























































































PLATE 26. 


This Plate exhibits several designs of Gothic roofs, the maimer of 
framing which is sufficiently indicated by the Plate. 

Fig. 1 is constructed entirely of wood. 

Fig. 2 of wood, strengthened with iron straps and bolts; and 

Fig. 3 with still less wood, but supported by iron rods; and, un • 
doubtedly, the strongest roof of the three. 

The first is, however, a neat, cheap, and very simple plan, and suf¬ 
ficiently strong for a roof having a steep pitch, and of not more than 
40 feet span. 

(87) 


I 


PLATE 27. 


Plate 27 represents two designs for church roofs, with arched or 
vaulted naves. 

In Fig. 1 the arch is formed of 2 inch planks, from 6 to 8 inches 
wide, after being wrought into the proper curve. These planks are 
doubled, so as to break joints, and firmly spiked together. Lighter 
arches, of similar construction, are sprung, at a distance of 16 inches 
apart, between the bents, for supporting the lathing. 

In Fig. 2 the arch is formed of 3 inch planks, 10 to 12 inches wide, 
and made in three sections, and spiked to the braces, as represented. 

Note .—The foregoing designs for roofs have been selected from more than a hundred 
drafts in the Author’s possession, and are believed to be the best selection ever offered 
to the public eye. The number could have been increased with ease to an indefinite ex¬ 
tent; but it has been deemed necessary to insert those only which are at once excellent 
aud practicable, and which combine the latest improvements. 

( 88 ) 


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PLATE 28, 


Plate 28 exhibits the frame-work of a church spire, 85 feet high 
above the tie beam, or cross timber of the roof. This is framed square 
as far as the top of the second section, above which it is octagonal. It 
will be found most convenient to frame and raise the square portion 
first; then to frame the octagonal portion, or spire proper, before rais¬ 
ing it: in the first place letting the feet of the 8 hip rafters of the spire, 
each of which is 48 feet long, rest upon the tie beam and joists of the 
main building. The top of the spire can, in that situation, be conve¬ 
niently finished and painted, after which it may be raised half way to 
its place, when the lower portion can be finished as far down as the 
top of the third section. The spire should then be ; aised and bolted 
to its place, by bolts at the top of the second section at AB, and also 
at the feet of the hip rafters at CD. The third section can then be 
built around the base of the £pire proper; or the spire can be finished, 
as such, to the top of the second section, dispensing with the third, 
just as the taste or ability of the parties shall determine. 

Fig. 2 presents a horizontal view of the top of the first section. 

Fig. 3 is a horizontal view of the top of the second section, after the 
spire is bolted to its place. 

The lateral braces in the spire are halved together, at their intersec¬ 
tion with each other, and beveled and spiked to the hip rafters at the 
ends. These braces may be dispensed with on a low spire. 

A conical finish can be given to the spire above the sections, by 
making the outside edges of the cross timbers circular. 

The bevels of the hip rafters are obtained in the usual manner for 
octagonal roofs, as described in Plate 20 

Note .—In most cases the side of an octagon is given as the basis of calculation in find¬ 
ing the width and other dimensions; but in spires like this, where the lower portion is 
square, we are required to find the side from a given width. The second section in this 
steeple, within which the octagonal spire is to be bolted, is supposed to be 12 feet square 
outside; and the posts being 8 inches square, the width of the octagon at the top of this 
section, as represented in Fig. 3, is 10 feet 8 inches, and its side is 4 feet 5.02 inches, 
as demonstrated in the explanation of the Table for Octagonal Roofs (No. 3). 

The side of any other octagon may be found from this by proportion, since all regular 
octagons are similar figures, and their sides are to each other as their widths, and, con¬ 
versely their widths are to each other as their sides.—See Explanation of Table No. 3. 

(89) 


Plate 29 exhibits the plan of a large dome of 60 01 75 feet span, 
built upon a strong circular stone or brick wall. In constructing this 
roof, there are four bents framed, like the one exhibited in Fig. 1, all 
intersecting each other beneath the king post at the centre. The tie 
beams in the first and second bents are of full length, and halved 
together; those in the third and fourth bents are in half lengths, and 
mitred to the intersection of the first and second. 

The King Post 

Has eight faces, and on each face two braces; one large bre^e from 
the top of the post to the end of the tie beam, and one small brace 
from the bottom of the post to the middle of the large brace. These 
four tie beams are supported by eight posts, extending from the top 
of the main wall to the ends of the beams, and sach one braced as 
represented in the figure. 

Two circular arches, constructed of planks, as described in Plate 
27, are then sprung, one above and one below each bent, as repre¬ 
sented in the figure. Between each of these four arches, three others 
are constructed, supported by short timbers, framed into the ties of 
the tie beam, as represented in Fig. 2. 

Fig. 2 is a horizontal section of the dome, drawn through it at the 
main tie beam AB, which corresponds with AB, in Fig. 1. 

Fig. 3 is a horizontal view of the apex of the dome, where all the 
32 arches intersect each other, showing the mode of beveling them at 
their intersection. 

(90) 



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PAET III. 



























































_ 



I 











Plate 30. 



Fig.3. 



Fig./. 



































































































































BRIDGE BUILDING. 




PLATE 80. 

* * 

STRAINING BEAM BRIDGES. 

Plate 30, Fig. 1 represents a straining beam bridge of 30 feet span, de¬ 
signed for t. common highway. The stringers or main timbers are 35 
feet long, extending over each abutment to a distance of 2| feet. The 
straining beam is equal in length to J of the span, or 10 feet. The 
supporting rods are 8 feet 2 inches long: 1 foot is allowed for the 
thickness of the stringer, 10 inches for the needle beam, and 4 inches 
nut head and washers; leaving 6 feet as the rise of the brace, or the 
distance of the top of the straining beam from the top of the stringer. 

The length of the brace can therefore be found, as usual, by extracting 
the square root of the sum of the squares of the run and the rise. 

Bevels. 

The bevel at the foot of the brace is like that at the foot of a rafter, 
and is obtained in the same manner. The bevel at the upper end of 
the brace and the bevel of the straining beam are equal to each other, 
and are each equal to half that of a rafter of the same rise and run. 

Fig. 3 exhibits the horizontal plan of the floor timbers, and the 
manner of laying both the joists and the planks. 

A moderate degree of camber should be given to every bridge of 
this kind, by screwing up the supporting rods. 

Bill of Timber, 

2 Stringers, 12 by 12, in. 35 feet long, Board measure =840 feet. 


4 Braces, 

8 by 10, 

it 

12 

u 

u 

u 

a 

=160 

ti 

2 Straining beams 

8 by 10, 

it 

10 

ft 

it 

It 

u 

=133 

it 

2 Wall plates, 

10 by 12, 

it 

16 

a 

it 

(l 

a 

=320 

a 

2 Needle beams, 

8 by 10, 

l( 

18 

it 

l 

ti 

u 

=240 

a 

6 Joists, 

3 by 10, 

u 

12 

a 

tt 

u 

it 

=150 

.< 


( 93 ) 



94 


CARPENTRY MADE EASY. 


6 Joists, 3 by 10, 22 feet long, Board measure=245 feet. 

932 feet, 2*inch planks “ “ =932 u 


Total timber, B. M. 3020 u 

Bill of Iron. 

4 Supporting rods, 1J in. diameter, 8 ft. 2 in. long, each 34 \ lbs.=138 lbs. 
8 Washers, 4 lbs. each; and 4 nuts 1 lb. each, = 36 “ 

4 Bolts, 1 in. diameter 22 in. long, each 5J lbs. = 22 “ 

8 Washers, 1 lb. each, and 4 nuts f lb. each, =11 “ 

40 lbs. spikes, = 40 “ 

247 “ 

Estimate of Cost. 

3020 feet lumber, @ $15 per M. =$45.30 

247 lbs. iron, @ 7c. “ lb. = 17.29 

Workmanship, @ $10 “ M. Board measure. = 30.20 

Total cost, $92.79 


Fig. 2. In respect to this bridge, it is only necessary to say that it is 
constructed upon the same principle as the former; the difference being 
caused only by the increase of the span, and this difference being suffi¬ 
ciently represented by the Plate. 

In raising the former of these bridges, no false work or temporary 
supports are needed, but for this one they may be. 

Concerning the economy and durability of these bridges, it may be 
proper to observe that they are comparatively simple and cheap; and 
they are also sufficiently strong, so long as the supports maintain their 
vertical position. But this plan has two objections. 

1. The absence of side braces induces a leaning or twisting of the 
braces, caused by their pressure toward each other; and when this 
twisting , or torsion as it is often called, has once commenced, it cannot 
well be remedied. It may, however, be guarded against, to a certain 
extent, by such a modification of the design as will allow of two sup¬ 
porting rods at each end of the needle beams—these rods being crossed; 
one passing inside of the stringer, and the other at some distance outside 
of it, toward the end of the needle beam. 

2. The absence of counter braces exposes the bridge to injury from 
vibration; which is specially destructive to the stone-work of the 
abutments, the repeated jars being almost sure to break the mortar 
and loosen the stones. The use of a wall-plate serves in some degree 
to obviate this objection; and in case of the bridge being supported by 
trestles, it disappears. 










fr'l/tar Plan 


Plate 31. 


o. 


d 


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L. 


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£ 


G 


n 




u 


0 


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$5 







Tlif ((.teonliardt & Smi.Phila 

























































































































































































PLATE 31. 


This bridge is more expensive and more durable than those before 
represented, as it is also less liable to the objections mentioned con¬ 
cerning them. The counter braces of this bridge are sufficient to pre¬ 
vent inj urious effects from vibrations; and the size of the posts, or up¬ 
right ties, when secured by straps of iron, as represented, will also 
prevent the torsion or twisting of the braces, to which the others are 
liable. The manner of framing this bridge is sufficiently indicated 
by the Plate; and the lengths and bevels of the braces are obtained as 
usual. 


Bill of Timber. 


2 String pieces, 

12 by 12 in. 65 feet long 

1560 feet , 

2 Straining beams, 

12 by 12 “ 18 ‘ 

a 

432 

M / 

4 Long braces, 

12 by 12 “ 26 ‘ 

u 

1248 

tl 

4 Short end braces, 

10 by 12 “ 15 ‘ 

a 

600 

it 

4 Middle braces, 

10 by 12 “ 10 ‘ 

tt 

400 

tt 

4 Counter braces, 

10 by 12 “ 9 ‘ 

tt 

360 

ti 

4 Long posts, 

12 by 12 “ 10 ‘ 

it 

480 

it 

2 Middle posts, 

12 by 12 “ 9 ‘ 

it 

216 

it 

4 Short posts, 

12 by 12. “ 6 ' 

u 

288 

it 

2 Wall plates, 

6 by 12 “ 18 

ti 

216 

it 

5 Needle beams, 

10 by 10 “ 20 ' 

tt 

833 

it 

12 Joists, 

3 by 10 u 24 

It 

720 

u 

6 Joists, 

3 by 10 “ 18 ‘ 

ft 

270 

tt 

2000 feet, B. M., ol floor plank, 16 ‘ 

It 

2000 

it 


* 


9623 

ti 


(95) 


/ 






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PLATE 32. 

This Plate presents a view «\f a bridge offered as an improvement 
of the Howe Bridge, of a moderate span, by shortening the upper 
chord, and bracing the ends of it in the same manner as in a straining 
beam bridge. In the Howe Bridge, the upper chord is of the same 
length as the lower one, and the braces and counter braces are placed 
in a uniform manner throughout the entire length. In the plan repre 
sented in this Plate, by reducing the length of the upper chord to the 
limit of a single piece of timber, it is proposed to secure, at least, an 
equal degree of strength to the ordinary Howe Bridge, and at the 
same time to effect economy in both material and labor. 

The ends of the braces are left square, and the proper bevels are 
made upon the angle blocks, which are of hard wood, or of cast iron, 
and are let into the chords to the depth of 1 inch or more. 

The main braces all lean inward toward the centre, of the span, and 
are double, passing one outside and one inside of the counter braces, 
which are single, leaning in the opposite direction from the centre to¬ 
ward the ends, each brace passing between each pair of main braces, 
and are all three bolted together at their intersection. 

The lower chords in each truss, or each side, are three in number, 
and bolted together in the most firm manner possible. Hard wood 
keys, 2 inches thick, 6 inches wide, and 12 inches long, are inserted 
on each side of every joint, and at certain intervals even where there 
are no joints. These keys are let into the chords .only about three 
fourths of an inch on each side, leaving a half inch space between the 
chords for the free circulation of air. 


Bill of Timber. 


2 Upper chord pieces, 

10 by 14 

in. 54 feet long=1260 feet. 

4 Long end braces, 

10 by 14 

“ 22 

it 

“ =1027 

<< 

4 Short “ “ 

6 by 6 

“ 12 

u 

“ = 144 

a 

4 Short end counter braces, 

4 by 6 

“ 12 

a 

“ = 96 

it 

32 Middle main braces, 

5 by 7 

“ 13 

a 

“ =1212 

a 

12 Counter braces, 

4 by 7 

“ 13 

a 

“ = 364 

a 

6 Lower chord pieces, 

6 by 12 

“ 31 

a 

“ =1116 

i1 

q a u u 

6 by 12 

“ 40 

a 

“ = 960 

u 


( 96 ) 




































































































































































































































































































































































RBIEGE BUILDING. 


97 


4 Lower chord pieces, 

6 by 12 

in. 30 feet Ion 

g= 720 feet. 

^ it it tt 

6 by 12 

“ 23 

it 

a 

= 552 

ft 

2 Wall plates, 

8 by 12 

“ 20 

tt 

tt 

= 320 

tt . 

32 Joists, 

4 by 12 

“ 18 

it 

tt 

=2304 

a 

10 Lateral braces, 

4 by 6 

“ 24 

it 

it 

= 480 

tt 

3000 feet, B. M., floor plank, 




3000 

a 


Bill of Iron 


it 


8 Next to middle 
16 End rods, 

8 Short end rods, 

4 Long cross rods, 

4 End bolts, 

24 Lower chord bolts, 
16 Brace bolts, 


it 


u 


18 Plates “ 

18 “ “ 

96 Washers, 

48 Nuts for small bolts, 


1 in. 

diam., 12 ft. 2 

in. 

long, 

32 lbs 

— 

128 lbs 

11 

tt 12 “ 2 

n 

it 

51 “ 


408 

11 

1J 

“ 12 “ 2 

it 

tt 

73 “ 

r= 

1168 

tt 

1J 

“ 6 “ 1 

it 

a 

36J“ 

— 

292 

ft 

1 

“ 18 “ 3 

tt 

tt 


— 

194 

<t 

1 

“ 2 “ 0 

it 

a 

5J“ 

— 

21 

1 1 

1 

“ 22 

tt 

tt 

5 “ 

= 

120 

11 

f 

“ 16 

it 

tt 

2 “ 

= 

32 

it 

rods, 




2 “ 

— 

144 

it 

“ i 

in. thick, 4 w., 

14 long, 

12 “ 

= 

216 

it 

tt 

a 4 w., 19 long, 

16 “ 

— 

288 

<t 





1 « 

= 

96 

tt 





1 “ 

— 

48 

it 


Note .—The cost of labor in constructing this bridge is estimated at .$11,00 per thcU' 
sand, B. M., of the timber required. 


7 



PLATE 33. 




TRESTLE BRIDGES. 

This Plate exhibits the design of a bridge supported from below; 
and, for a moderate span, it is one in which the important elements 
of simplicity, strength, and durability, are well combined. 

The plan of this bridge is so simple, as to require little further ex¬ 
planation than the inspection of the Plate. It will be perceived that 
the bearings are K) feet apart, and that the braces are framed to cor¬ 
respond. The cn <ss timbers are extended out several feet on each 
side, to give room for bracing the hand-rail. 

This bridge is supported by trestles; and the Plate represents the 
manner of framing the end ones and the middle one. It is of the 
utmost importance that the embankments behind the end trestles are 
perfectly solid, as on their firmness depends the whole strength of the 
bridge. 


Bill of Timber for One Span. 


4 String pieces, 

12 by 12 

in., 28 feet 1 

ong, 

=1344 feet. 

2 Straining beams, 

12 by 16 

“ 30 

a 

“ . 

= 960 

a 

4 Long braces, 

10 by 10 

“ 24 

ti 

a 

= 800 

it 

4 Short braces, 

10 by 10 

“ 14 

u 

a 

= 467 

u 

6 Cross timbers, 

6 by 12 

“ 24 

tt 

i 

= 864 

u 

12 Hand-rail posts, 

6 by 6 

“ 4J 

a 

u 

= 162 

<( 

104 feet (lineal) hand railing, 

6 by 6 

u 



= 312 

a 

12 Hand-rail braces, 

3 by 4 

“ 4J 

u 

u 

= 54 

tt 

12 Joists, 

3 by 10 

“ 20 

tt 

it 

= 600 

u 

6 

3 by 10 

« 11 

It 

u 

= 165 

u 

1660 feet, B. M., floor plank, 

• 




=1660 

ti 

Total 





7388 

u 

Bill of Timber for the Two End Trestles. 



4 Posts, 

12 by 12 

in., 13 feet long, 

= 624 feet 

8 Studs, 

6 by 12 

“ 13 

u 

it 

= 624 

u 

4 “ 

6 by 12 

“ 8 

It 

i 

= 192 

< 


( 98 ) 
























































































































































































































































































































































* 

































- 

r 



















BRIDGE BUILDING. 


99 


4 St ads, 

2 Mud sills, 

2 Caps, 

1000 ft., 2 in. liard wood 


6 by 12 in., 6 feet long, = 144 feet. 
12 by 12 “ 52 “ “ =1248 “ 

12 by 12 " 20 “ “ = 480 “ 

plank for supporting ernbankm’t,=1000 a 


4312 “ 


Bill of Timber for Middle Trestle. 


3 Posts, 

12 by 12 

in., 

13 feet long, 

B.M., = 468 feet, 

1 Mud sill, 

12 by 12 

a 

30 “ 

u 

“ = 360 

u 

1 Cap, 

12 by 12 

u 

20 “ 

u 

“ = 240 

LI 

4 Post bead braces, 

4 by 4 

u 

5| “ 

it 

= 30 

tl 

2 “ foot braces, 

8 by 8 

u 

8 “ 

(( 

“ = 90 

a 






1188 

a 

For tbe two end trestles. 

i 




4312 

a 

Total of tbe three trestles, Board measure, 


6500 

te 





PLATES 34 & 35. 

Plates 34 and 35 represent a strong trestle bridge, such as is often 
used for rail-roads in crossing small streams and ravines, where the 
banks are high, and where there is little danger from ice. The Author 
of this work has constructed bridges of this kind at Spring Creek, Bu¬ 
reau Co., and at Nettle Creek, Grundy Co., on the Chicago and Bock 
Island Eail-road; and one on the plank road, between Peru and La 
Salle, in La Salle Co., Ill.—the last with posts, 51 feet high. 

In framing the trestles , the posts are framed into the sills and caps as 
usual; but the braces are bolted upon the outside with inch bolts. The 
outside lower braces of the trestles, marked in the plan C, C, have 1 
foot run to 2 feet rise; the posts of the trestles at A are set in such a 
manner as to act as braces, having 1 foot run to 4 feet rise. The hori¬ 
zontal lateral braces are also laid and bolted between the longitudinal 
timbers and cross timbers, without being framed into them. The 
lower longitudinal timbers are let into the posts to the depth of 2 
inches, and lapped across the posts, one on one side, and the other on 
the other side, where they are bolted to the posts and to each other. 

The bearings are ten feet apart , and each bearing is supported either 
by a post or a brace; these braces are framed to a 10 feet rise and a 9 
feet run, and the upper ends are bolted to the longitudinal timbers, as 
represented in the Plate. 

A bill of timber and iron, which is here subjoined, will assist the me¬ 
chanic in framing a bridge of this kind more than any extended de¬ 
scription could do. (The small letters in the Plate refer to the bolts.) 


Bill of Timber. 


6 Sills, 

12 by 12 

in., 

33 feet long, 

B. M., =2376 feet 

1 Sill, 

12 by 12 

u 

18 “ 

u 

“ = 216 “ 

1 “ 

12 by 12 

li 

16 « 

It 

“ = 192 “ 

18 Posts, 

12 by 12 

It 

25 “ 

u 

“ =5400 “ 

3 “ 

12 by 12 

a 

12 “ 

u 

“ = 432 “ 

3 “ 

12 by 12 

u 

10 “ 

it 

“ = 360 “ 

6 Caps, 

12 by 12 

a 

19 “ 

(( 

“ =1368 “ 

2 “ 

12 by 12 

u 

11 “ 

It 

“ = 264 “ 

3 Lower cross timbers, 

6 by 12 

a 

19 “ 

u 

“ = 342 “ 


( 100 ) 


< c 
t t c 





































































. 








































. 

. 


* 

' 





























































































































































Floor Flans. 



Tiif'i. r.iarh.rtlt ^ .■imi.T’tula. 










































































































































































































































































































































BRIDG3 BUILDING, 


101 


6 Lower longitudinal timbers, 6 by 12 in., 

35 feet long, 

=1260 feet. 

18 Braces, 

8 by 10 “ 

14 

a it 

=1680 

u 

6 “ 

8 by 10 « 

15 

a it 

= 600 

(( 

6 String pieces, 

9 by 12 u 

27 

a it 

=1458 

it 

3 “ « 

9 by 12 “ 

32 

a it 

= 864 

it 

9 Cross timbers, 

10 by 12 “ 

19 

a a 

=1710 

it 

16 Lateral braces, 

4 by 8 “ 

27 

a a 

=1152 

It 

8 Kail stringers, 

12 by 14 “ 

30 

a a 

=3360 

a 

4 Bolsters, 

12 by 12 “ 

6 

n a 

= 288 

tt 

6 Cross braces, 

4 bv 8 “ 

14 

a a 

= 224 

a 

6 “ “ 

4 by 8 “ 

19 

a a 

= 304 

n 

2 “ “ 

4 by 8 “ 

16 

a n 

= 85 

it 

2 “ “ 

4 by 8 a 

18 

1 1 a 

= 96 

it 

Total, Board measure, 



24,031 

a 


Bill of Iron. 





6 Bolts (letter a), 

32 in. long, 

1 

in. diam., 

= 42 J lbs 

12 “ ( “ b\ 

36 

u 

n 

— 95J 

a 

12 “ ( “ c), 

18 “ 

a 

u 

= 24 

a 

18 “ ( “ d), 

22 “ 

u 

a 

= 87} 

a 

3 " ( “ e), 

22 “ 

it 

a 

= 15 

it 

to 

•> 

31 “ 

a 

a 

=185 

tt 

18 “ ( “ ' g), 

26 u 

a 

it 

=103} 

it 

8 “ ( “ A), 

28 

a 

n 

= 50 

tt 

294 Heads, nuts, and wasters, @ 1 lb. each, 


=294 

it 

18 Bolts (letter i), 

20 in. long, 

3 

4 

in. diam., 

= 45 

it 

6 “ ( “ j), 

11 

it 

a 

= 8} 

tt 

20 “ ( “ k), 

18 “ 

it 

it 

= 45 

a 

2 “ ( “ 0, 

22 u 

a 

a 

= 5} 

a 

138 Bolt-beads, nuts, 

and washers, @ } lb. each, 

=103 

it 


Total of iron, 1100J “ 




PLATE 36. 


ARCHED TRUSS BRIDGES. 

This Plate represents a design of a Burr Bridge without counter 
braces, but combined with an arch beam. This mode of construction 
is designed either for railroad bridges, or for common road bridges of 
a great span. If wanted for a common road, and the span be not 
more than 150 feet, the arch beam may be safely dispensed with; and 
in that case, counter braces should be introduced; but if the bridge 
be designed for a railroad, the arch beam should never be omitted. 

The panels of bridges of this kind ought never to be as great in 
extension as in height between chords; or, in other words, the rise of 
the braces should always be greater than their run; and practically, 
it is expensive and inconvenient to extend the panels more than 12 
or 14 feet. In all bridges of this kind, the greatest strain upon the 
braces is at the end of the span; and it will be most proper to use the 
best and largest pieces of timber for the end braces, and those of in 
ferior quality, if such must be used somewhere, should be placed in 
the middle. 

The posts should be sized down at the lower end, where they pass 
through the lower chord, to about 6 inches in thickness; the chord 
pieces should also be cut out to the depth of 1 inch on each side of 
the post, and both locked into the post in the firmest possible manner, 
in order to resist the thrust of the brace. The post should also be 
boxed into the upper chord not less than 1 inch. 

In scarfing the lower chord pieces, they must be so arranged that 
only one splice be made at the same place; and if the bolts which pass 
through the scarfing extend also through both lower chord pieces, 
(the short piece inserted to lock the joint being of just sufficient thick¬ 
ness to fill the space between the two chord pieces), it would be still 
better than that plan represented in the Plate. 

It will be found necessary, in a bridge of this kind, to make the 
main braces at least 1J inches longer than the exact calculation would 
require, in order to produce the necessary camber, and to guard 
against the settling of the centre of the span below the general level, 
which will be likely tc happen if not guarded against, from the com* 
( 102 ) 


♦ 



ITiC'i I Punluirdt&Son i'luia. 




























































































































































































































































































































































BRIDGE BUILDING. 


103 


pression and shrinkage of the timber, and which would materially 
weaken the bridge ; and whatever camber the bridge is designed to 
have, must be given to it on its first erection, before the false works 
are removed, since the camber cannot afterward be increased as it can 
be in most of the bridges represented on the preceding Plates, where 
supporting rods, in those plans, occupy the place of the posts in this. 

For floor plan, see Plate 87. 

Bill of Timber for One Span. 


2 Wall plates, 

10 by 12 

in.. 

, 22 feet long, 

B. M.= 440 feet. 

8 Lower chord pieces, 

7 by 14 

it 

34 

tt 

it 

“ = 2286 

u 

g tt tt u 

7 by 14 

it 

43 

tt 

it 

“ = 2809 

a 

4 Upper “ “ 

11 by 11 

u 

36 

it 

it 

“ = 1452 

tt 

£ It it It 

11 by 11 

a 

44 

u 

it 

“ = 1774 

tt 

46 Floor beams, 

4 by 12 

tt 

20 

n 

it 

“ = 3680 

tt 

60 Arch pieces, 

7 by 10 

a 

33 

tt 

it 

“ =11550 

it 

12 Queen posts, 

11 by 15 

u 

19 

a 

it 

“ = 3135 

tt 

8 “ “ 

11 by 14 

it 

19 

tt 

it 

“ = 1950 

a 

8 “ “ 

11 by 13 

it 

19 

it 

a 

“ = 1811 

a 

2 Centre posts, 

11 by 16 

tt 

19 

it 

tt 

“ = 557 

a 

4 Arch braces, 

12 by 14 

it 

18 

it 

a 

“ = 1008 

it 

28 Panel girths, 

6 by 8 

it 

11 

it 

tt 

“ = 1232 

a 

12 Scarfing blocks, 

6 by 14 

it 

5 

a 

a 

“ = 420 

tt 

16 Tie beams, 

10 by 10 

a 

19 

tt 

tt 

u = 2533 

it 

32 Knee braces, 

4 by 6 

It 

6 

tt 

a 

* “ = 384 

tt 

18 Lateral braces, 

4 by 8 

tt 

24 

it 

tt 

“ = 1152 

u 

7000 feet 8 inch floor plank, 


14 

tt 

it 

= 7000 

a 


Bill of Iron. 





9 Cross rols, 

1 in. diam. 

19 feet 3 

inches long =460 lbs. 

22 Arch bolts, 

1 « 


2 

“ 9 

a 

“ =160 

tt 

24 Splice bolts, 

1 “ 


1 foot 8 

it 

^ =106 

tt 

80 Post bolts, 

1 “ 


1 

“ 8 

tt 

“ =132 

tt 

60 Arch bolts, 

3 tt 


2 feet 0 

a 

“ =179 

tt 

Head, nut, and washer to 

each bolt, 

1 lb. each, 


345 

u 


PLATE 37. 




This Plate represents a first-class Howe Bridge for a Railroad, and 
as one of similar principles and construction has been fully explained in 
the description of Plate 32, page 96, it will only be necessary here to 
add the bills of timber and iron, in detail. 

For a common highway, a bridge of this length will admit of some 
modifications, as it is heavier than necessary. 

The cost of a bridge of this kind, in the State of Illinois, would be 
about $30 per lineal foot. 


Bill of Timber for One Span Howe Bridge, 150 feet long. 


4 

Wall plates, 

10 by 12 

in., 

20 feet long, 

= 800 feet. 

8 

Bolsters, 


10 by 14 

tt 

10 

a 

tt 

= 933 

tt 

12 

Lower chords, 

6 by 14 

tt 

42 

tt 

tt 

=3528 

tt 

2 

tt 

tt 

6 by 14 

tt 

32 

tt 

tt 

=» 448 

tt 

2 

tt 

' a 

6 by 14 

tt 

22 

tt 

tt 

= 308 

tt 

2 

tt 

tt 

6 by 14 

a 

12 

tt 

tt 

= 168 

tt 

12 

tt 

tt 

7 by 14 

tt 

42 

a 

tt 

=4116 

tt 

2 

tt 

tt 

7 by 14 

tt 

32 

tt 

tt 

= 522 

u 

2 

tt 

(( 

7 by 14 

tt 

22 

tt 

tt 

= 359 

a 

2 

tt 

tt 

7 by 14 

tt 

12 

tt 

tt 

= 196 

tt 

12 

Upper 

a 

6 by 12 

tt 

42 

tt 

tt 

=3024 

tt 

2 

(( 

tt 

6 by 12 

tt 

32 

tt 

tt 

= 384 

tt 

2 

tt 

tt . 

6 by 12 

tt 

22 

tt 

tt 

= 264 

tt 

2 

tt 

a 

6 by 12 

tt 

12 

tt 

tt 

= 144 

n 

12 

tt 

tt 

7 by 12 

tt 

42 

tt 

tt 

=3328 

tt 

2 

tt 

tt 

7 by 12 

tt 

32 

tt 

tt 

= 448 

tt 

2 

a 

tt 

7 by 12 

a 

22 

tt 

tt 

= 308 

tt 

2 

tt 

tt 

7 by 12 

tt 

12 

tt 

tt 

= 168 

tt 

8 

Main braces, 

10 by 11 

tt 

24 

'tt 

tt 

=1760 

tt 

8 

a 

tt 

10 by 10 

tt 

24 

tt 

tt 

=1600 

a 

8 

tt 

tt 

10 by 9 

tt 

24 

tt 

tt 

=1440 

tt 

8 

tt 

tt 

10 by 8 

tt 

24 

tt 

tt 

=1280 

tt 

8 

it 

tt 

10 by 7 

tt 

24 

tt 

tt 

=1120 

tt 

16 

tt 

tt 

10 by 6 

a 

24 

tt 

tt 

=1920 

a 

28 

Counter 

tt 

6 by 8 

tt 

24 

tt 

tt 

=2688 

tt 

56 

Lateral 

a 

6 by 6 

tt 

18 

tt 

tt 

=3024 

u 

46 

Cross floor timbers, 

7 by 14 

tt 

20 

tt 

tt 

=7513 

tt 

12 

Rail stringers, 

10 by 12 

tt 

26 

tt 

tt 

=3120 

tt 

4 

Struts, 


5 by 12 

n 

16 

tt 

tt 

= 320 

tt 

16 

End supports, 

5 by 12 

tt 

20 

it 

tt 

=1600 

it 


( 104 ) 


















































































































































































































































, 





















































































BRIDGE BUILDING. 


105 


700 ft. Oak for clamps, 3 by 14 in., 12 feet long, = 700 feet. 

500 “ “ Packing blocks, 3 by 14 “ 12 “ “ = 500 “ 

500 “ “ Upper “ “ 3 by 12 " 12 “ “ = 500 u 

48,531 


Bill of Wrought Iron. 


8 

Supporting rods, 

If in. diam., 

22 feet 

8 

in. 

long, 

=1961 

lbs. 

40 

a a 

n 

a 

22 

a 

8 

a 

a 

=5436 

u 

24 

it it 

is 

a 

22 

a 

8 

a 

u 

=2720 

a 

24 

a a 

n 

a 

22 

a 

8 

u 

a 

=2310 

a 

6 

u a 

i 

a 

22 

a 

8 

u 

a 

= 380 

a 

8 

a u 

i 

a 

23 

a 

6 

a 

a 

= 517 

a 

30 

Lateral u 

i 

a 

19 

a 

6 

u 

a 

=1580 

a 

180 

Chord bolts, 

i 

a 

2 

u 

8 

a 

a 

= 720 

a 

24 

Gibs, 5 holes, 

1 by 6 in., 

3 

a 

0 

a 

a 

=1440 

a 

24 

“ 3 “ 

1 by 5J « 

2 

a 

6 

a 

a 

=1080 

u 

4 

“ 3 “ 

j by 5 “ 

2 

a 

6 

a 

a 

= 127 

tt 

566 

lbs. Nuts, 








= 566 

a 










18,837 





Bill of Castings. 






4 

Half angle blocks, 



94 

lbs, 

, each, 

= 376 

lbs. 

24 

Angle blocks, 




178 



u 

=4272 

a 

24 

a a 




148 



u 

=3552 

a 

4 

a a 




132 



a 

= 528 

a 

60 

Lateral angle blocks, 



21 



a 

=1260 

a 

136 

Washers, 


4 in. diam, 

3 



a 

= 408 

a 

360 

Chord bolt washers, 

3 “ 


H 


u 

= 450 

a 










10,846 



1st Set of Pods, next to abutments, 2 rods, If in. iron, 3 rods, 1J in. iron. 


2d 

a 

a 

a 

a 

2 

a 

1} 

a 

3 “ If “ 

3d 

a 

a 

a 

a 

2 

a 

n 

a 

3 “ If “ 

4th 

a 

a 

a 

a 

3 

a 

u 

a 


5th 

a 

a 

a 

a 

3 

a 

it 

u 


6th 

a 

a 

a 

a 

3 

a 

u 

a 


Center “ 

a 

a 

a 

3 

a 

i 

a 



Note .—It will be seen that the first 3 panels, next to the abutments, are supported by 5 rods 
each, and that the outside and inside rods are beyond the chords, passing through tubes in the 
angle-blocks, also the heaviest main braces are used next to the abutments, growing lighter as 
they approach the center. 





PLATE 38. 


Tliis bridge is similar, in its general principles of construction, to 
the one represented in Plate 37; but is quite different in its minor de¬ 
tails, being much heavier and stronger, as well as more expensive. 
The main differences are these: Counter braces are employed in this 
bridge, which are omitted in the other; this has two sets of posts and 
main braces, and but one arch beam to each truss, while the other 
bridge has two arch beams and one set of posts and braces; the chord 
pieces in this bridge, instead of being placed side by side, with their 
edges vertical, with an open space between them for the circulation 
of air, are placed one upon the other, with their edges horizontal, and 
their surfaces in close contact. 

The upper chord is in three sections, and the lower chord and the 
arch beam are each in four sections; each chord piece and arch piece 
being 6 inches deep and 12 inches wide; making the combined upper 
chord 12 by 18 inches, and the combined lower chord and arch beam 
each 12 by 24 inches. 

The foot of the arch beam rests upon a cast iron shoe, secured by 
iron straps to each of the lower chord pieces ; each shoe having four 
flanges, and each flange beveled to fit the square end of each section 
of the arch beam. 

There is one set of counter braces to each truss, each counter brace 
passing between each pair of main braces, to which it is bolted at their 
intersection. The foot of each counter brace rests upon an angle block 
fixed upon the lower chord, at the foot of each pair of posts, and the 
upper end of each counter brace rests against the arch beam at its inter¬ 
section with the next pair of posts. A key is inserted, however, be¬ 
tween the upper end of each counter brace and the arch beam, by means 
of which the whole structure can be kept tight, and the relative strain 
upon the arch beam and the chords can, to some extent, be regulated 
and proportioned. Each pair of posts is bolted together with four 
bolts—one above, and one below each chord. 

Bridges of this style are in extensive use on the New York and 
Erie Bail-road, where they have been proved to be of great strength 
and stability. 

( 106 ) 



38 





Tl. eo. Lfiin’ianl t X Son. I'Ki la. 








































































































































































































































































































































































































GENERAL PRINCIPLES OF BRIDGE BUILDING. 


In concluding this Part of the work, it is proper to bring together 
into one place the most important principles and most useful hints to 
practical builders, which we have been able to gather, either from the 
study of other works,* or from the lessons of our own experience. 

Size of Timber and Iron required to enable a Bridge of a given Span to 

sustain a given Load. 

The most proper way of ascertaining the resisting powers of timber 
and iron is by actual experiment; and it has been found by such ex 
periment, that the greatest safe strain for sound timber is about 1,000 
lbs. per square inch, measured on the square end of the timber, the 
strain being one of either extension or compression, but applied in the 
direction of the grain of the wood. It has also been ascertained by 
experiment, that the greatest safe tensile strain , as it is technically 
called, that is the lifting or supporting strain, of large wrought iron 
rods, is 10,000 lbs. per square inch. Small wire or nail rods, manu 
factured with more care, and of the best materials, can, undoubtedly, 
sustain a much greater weight than this. 

In proportioning the different parts of a bridge, however, it is cus¬ 
tomary and expedient to allow a considerable excess of strength in 
favor of stability. The deterioration of timber, caused by age, must 
be taken into the account; for after a wooden bridge has been in use 
for some years, it becomes much weaker than when first erected. 

The weight of the bridge itself must also be considered in deter¬ 
mining the load which it is able to sustain, and this weight it is con¬ 
sidered safe to assume at 35 lbs. to the cubic foot of timber employed 
If the quantity of timber in a given bridge is equal to 30 cubic feet 
to every foot in length, as is asserted by Ilaupt to be the case with 
the average of the Howe Bridges on the Pennsylvania Eail-road, then 
the weight of the structure would be 1050 lbs. per lineal foot, or a 


* Many of these remarks are condensed and simplified from the work on “ Bridge 
Construction,” by Herman Haupt, A. M.—D. Appleton & Co., New York, a work more 
especially designed for the use of engineers than for practical builders, yet one which 
we commend to all persons interested in this part of Carpentry. 


( 107 ) 



108 


CARPENTRY MADE EASY. 


little more than half a ton per foot for the weight of the timber, ex¬ 
clusive of the iron. 

The greatest load that can be brought upon a rail-road bridge, 
with a single track, is when several locomotive engines of the first 
class, weighing about one ton per foot in length, are attached to¬ 
gether. So that the greatest strain upon such a bridge, including both 
its own weight and the weight of the load, is a little more than a ton 
and a half per foot. What, then, must be the dimensions of the tim¬ 
ber to resist this strain ? 

The Strain upon the Chords. 

When a beam is supported at the ends and loaded in tne middle 
until it breaks, it is observed that the fibres in the lower portion of 
the fracture are broken by being extended or pulled violently apart, 
and that those on the upper portion are broken by being compressed 
or jammed violently together. In theory, this compression is said to 
be equal to the expansion; that is, that it will require an equal force 
to tear the fibres apart as to break them by forcing them together, 
and the neutral axis in the beam, or the line where there is neither 
sufficient expansion nor compression to break the fibres of the timber, 
is said to be in the middle of the beam. But it is doubtful whether 
facts will warrant this conclusion. Common observation w r ould lead 
most persons to the opinion that timber has a greater power to resist 
compression than it has to resist expansion, and to this opinion we 
are ourselves inclined; but for the present purposes it will be suffi¬ 
ciently accurate to be governed by the theory usually adopted by en¬ 
gineers, as stated. 

The power of a bridge to sustain a load, and to resist the various 
strains upon it, may be compared to that of the beam supported at the 
ends—the strain on the upper chord being one of compression, and 
that on the lower chord one of extension ; and the strain on both being 
greatest in the middle of the span, and diminishing toward the ends. 
When the beam is laid over several supports, its strength for a given 
interval is much greater than when simply supported at the ends. 
The same principle is applicable to bridges; and when several spans 
occur in succession, it is of great advantage to continue the upper and 
lower chords across the piers. 

The greatest strain on the upper chord being in the middle of the 
span, is equal to that force which, being applied horizontally, would 
sustain one half the span with its load were the other half to be re 


BRIDGE BUILDING. 


109 


moved. In order to ascertain this force, multiply half the span and 
its load by one fourth its length, and divide that product by its height, 
measured from centre to centre of the upper and lower chords. 

For example, if the length of a span be 160 feet, and the height of 
the truss be 16 feet from centre to centre of upper and lower chords, 
and the weight of the loaded bridge be 1J tons to the lineal foot, the 
greatest strain upon the upper chord would be expressed by the 
product of 120 tons multiplied by 40, and the product divided by 16; 
which gives 300 tons, or 600,000 lbs. as the result. The reason of 
multiplying the weight of half the loaded span by 40 is, because 40 
feet is the middle of the half-span, or its centre of gravity; and the 
reason for dividing its product by 16 is, because that is the width of 
the truss; and the wider the truss, the greater leverage there is, and 
the less strain, for the same reason that a thick beam is stronger than 
a flat one, as there is less strain on the upper and lower surfaces of 
the thick beam from the same weight than in a flat one. Then, as 
each square inch is able to resist 1000 lbs., there must be 600 square 
inches in the end section of the upper chords, in order to enable them 
to sustain the weight required, or 300 inches in the upper chord of 
each truss. If, therefore, each chord is 12 inches deep, it must be 25 
inches wide; hence, three chord pieces, 12 by 8J inches, will contain 
the requisite material 

The strain on the lower chord is at least equal to that on the upper 
one; but the timbers being in several pieces, and the strain being one 
of extension, the joints are opened, and the whole strength of the tim¬ 
ber is not available; while in the upper chord the strain is one of com¬ 
pression, and the joints being pressed together, causes no loss to the 
resisting force of the timber. There must, therefore, be an additional 
quantity of timber in the lower chord; and each piece should be suffi¬ 
ciently long to extend through four panels, so that there can be three 
whole timbers and a joint in each panel. 

From the same data, similar calculations can easily be made for 
estimating the strain and fixing the dimensions of the other timbers. 


PART IV 


I 














* 



























































































’ 







































































PLATE 39. 


This Plate is introduced here to demonstrate more fully the proper 
way of obtaining the backing on hip rafters: continued from page 71, 
where we say that “ this is found on the square by taking the length of 
the hip rafter on the. blade, and the rise of the roof on the tongue; the 
bevel on the tongue is the backing required.” To find the proportional 
length of the hip rafter, we take a two foot rule and measure the 
diagonal distance on a square, by placing one end of the rule upon the 
17 inch mark of the blade, and the other end upon that mark of the 
tongue which corresponds with the rise of the roof to the foot; this gives 
the proportional length of the hip rafter; then apply this length to the 
blade of a square and the rise of the roof to the tongue, as above stated 
on page 71, and the true bevel will be obtained. 

To illustrate and make it more plain, we will find the backing of a 
hip rafter on a roof which rises 5 inches to the foot. On applying the 
two foot rule to the square at the 17 inch mark on the blade, and at the 
5 inch mark on the tongue, we find the diagonal distance to be 171- 
inches ; taking this last measure on the blade and the 5 inch mark on 
the tongue and applying them to a straight edge, the tongue will indi¬ 
cate the backing required, as shown at A, Fig. 1. 

The intersection of the two squares, each set in the same manner as 
above described, shows the backing on both sides of the rafter at B, and 
a bevel square set to the proper angle is exhibited at C. 

If the rise of the roof is 4 inches to the foot, we proceed as before, 
and find the diagonal or proportional length of the hip rafter to be 17§ 
inches, which we apply to the straight edge on the blade and the 4 inch 
mark on the tongue, and find the proper backing for a 4 inch pitch. In 
a similar manner a 10 inch pitch gives a diagonal of 19f inches, and in 
fine, this rule is correct for any pitch. 

The reason why we use the 17 inch mark on the blade of the square 
for this purpose is because the diagonal of a square foot is 16 r % inches, 
to which 17 inches is as near as we can practically work. See pages 132 
and 133, where these principles are fully explained. 

Fig. 2 exhibits a practical method of working off the corners of square 
timber to make it octagonal or eight square. In this Fig. TM represents 
any piece of timber not more than ’2 feet square, on which we place the 

( 111 ) 


112 


CARPENTRY MADE EASY. 


square as represented, with the heel at one edge and the end of the blade 
at the other; then prick the timber at the 7 and 17 inch marks accur¬ 
ately, and set the gauge accordingly and gauge each face, then take off 
the corners to these gauge marks. This method is sufficiently accurate 
for common purposes, the error being only of an inch, for the square 
of 7 is 49, and twice 49 is 98, and this is the face of the corner which is 
taken off, the middle face which is left being 10 inches, of which the 
square is 100. Should the timber be larger than 24 inches square, take 
2 ? of the width of each face, and proceed as before. 

Fig. 3 exhibits another method. Let AB represent the diagonal 
drawn across the end of a square stick of timber, and let the dotted lines 
intersecting at the center be arcs drawn from each corner with a radius 
of half the diagonal, then take off the corners to the points where the 
arcs cut the faces, and the octagon is exact. 

Mechanics, as a rule, square their work by the numbers 6, 8, and 10, 
which is correct, yet only a few of them know the reason why it is so. 
It is simply because these numbers represent the three sides of a right- 
angled triangle of which 10 is the hypotenuse, the square of which is 
equal to the sum of the squares of the other two sides; i. e., 6 times 6 is 
36, and 8 times 8 is 64, and the sum of 36 and 64 is 100, which is also 
the square of 10. 


























* 














































































PLATE 40. 


This Plate is introduced here to show the proper method ol bracing 
between posts that stand battering or inclined, as in a derrick for a wind¬ 
mill. In Fig. 1 the posts incline or batter two inches to the foot in rise; 
the. bevel of the posts at the top and bottom is obtained as usual, by 
taking 12 inches on the blade of the square, and 2 inches on the tongue, 
and also for the ends of the girder, which has the same bevel. 

The braces on the lower side of the girder must be set as many inches 
nearer the shoulder of the girder as the post gains in battering. If the 
post batters two inches in rising one foot, and if the brace is framed on 
a regular run of three feet, the toe of the brace will be six inches nearer 
the shoulder than it would be were the post to stand plumb. Hence the 
true distance of the toe of the lower brace from the shoulder of the 
girder is 2 feet 6 inches, and on the upper side the brace will be 3 feet 
6 inches, for a similar reason. 

To find the distance on the post for the toe of the brace, turn to the 
common rafter table, page 130, and take the length of a rafter for a 
building which is double the width of the run of the brace, and that will 
be the true distance from the girder to the toe of the brace either above 
or below the girder. For example, in Fig. 1 the braces on the girder 
are each framed for a regular run of 3 feet and the posts battering 2 
inches, we turn to the rafter table and find that, for a building 6 feet 
wide and with a rise of 2 inches to the foot, the length of the rafter is 
3 feet i 4 o 9 {j of an inch, which is the exact distance of the toe of the brace 
from the girder. 

Practically this distance may be found by the square, thus : Apply the 
square to the post, placing the end of the blade at the shoulder of the 
girder and the 4 inch mark on the tongue either above or below; then 
at this latter point apply the 12 inch mark of the blade again and the 
2 inch mark on the tongue still higher or lower upon the post, and this 
last point will be the true place for the toe of the brace, and on meas¬ 
uring the distance, it will be found to be 3 feet of an inch, as before. 

Should the battering or inclination of the posts be either more or less 
than 2 inches to the foot, then the same principles will apply, by vary¬ 
ing the use of the tables and the square accordingly. 

S (113) 


114 


CARPENTRY MADE EASY. 


Bevels of the Braces. 

There are three different bevels of the braces: 1. Those upon the 
girder are all regular and at an angle of 45°. 2. Those above the girder 
are as much more than 45° as the girder is out of square; and 3. Those 
below the girder are as much less than 45° as those above are more, the 
one being the complement of the other. The most convenient way to 
find these bevels is first to cut the regular bevels at the lower end of the 
braces, as at D in Fig. 2, then measure its length from D to E, then 
mark or cut the upper end at the same bevel parallel to the first, and 
then apply to this mark or section a bevel square which has been set to 
the exact bevel of the end of the girder; then one end of the blade will 
show the bevel of the lower brace, as at E, and by turning the bevel 
square upside down, the other end of the blade will show the bevel of 
the upper brace, as at B, Fig. 3. 

Height of the Girder. 

To ascertain the slant height of the girder from the sill upon the 
posts, suppose we want the perpendicular height to be 10 feet, then from 
the common rafter table as before, we find the length of a rafter for a 
building 20 feet wide, with a rise of 2 inches to the foot. This is found 
to be 10 feet lyVo inches, which is the slant height of the girder from 
the sill. Or take 5 lengths of the blade of the square, applied as before, 
with the 4 inch mark of the tongue upon the post, and the last point 
will be the true distance of the girder upon the post from the sill. 

Length of the Girder and Cap. 

To find the length of the girder, multiply its perpendicular height in 
feet, by the battering of both the posts to a foot, in inches; the product 
will be in inches to be subtracted from the length of the sill between the 
posts. In this example we subtract 40 inches from the inner length of 
the sill, because 10 feet, the height of the girder, multiplied by 4 inches, 
the battering of both the posts to a foot, produces 40 inches. 

In the same manner find the length of the cap by subtracting the 
product of its height, multiplied by the battering of the two posts, from 
the outer length of the sill. 

The Long Braces. 

To frame the long braces between the posts, as shown in Fig. 1, first 
find the distance between the posts at the bottom of the braces in the 
same manner as before in finding the length of the girder, then as the 
braces should be framed on a regular run, or at an angle of 45°, we see 


BATTERING POSTS. 


115 


that if the posts were 7 feet apart at the dotted line AA', the brace 
would rise just 6 feet and run 6 feet at its intersection with the post at 
B, because the point B is perpendicular to P, which is just 1 foot from 
A, for the post batters one foot in rising 6 feet, and we find the length 
of the brace from the brace table, page 140, to be 8 feet 5 T 8 o 2 o inches. 

If the distance AA' be either more or less than 7 feet, say 9 feet, then 
the line A'P may be found by proportion, thus: 

7 : 6 :: 9 : the required distance A'P. 

This example is here wrought out as follows: 

9 7: 6:: 9: 7.714 

_6 

7)54_ 

7.714 
_12 

8.568 Ans. 7 feet 8-,% inches. 

As this answer is a fractional number, the length of the brace cannot 
be found from the brace table, but must be obtained by calculation, as is 
explained in the introduction to that table, by reducing the number to 
inches, squaring it and doubling the square, and then extracting the 
square root from the sum, thus: 7 feet 8 t 5 q% inches = 92.56 inches, and 
92.56 X 92.56 — 8567.3536, double this and we have the following 
number, 17134.7072, from which we extract the square root, thus: 

17134.7072(130.89 Ans., 

1 


23)71 or 

69 

2608)234.70 10 feet 10 T 8 ^ inches, 

208.64 

26.0672 true length of brace. 

23.5521 

2.5151 

The bevels of the long upper braces will be the same as those of the 
shorter braces already found; i. e ., all the upper end bevels marked B 
and B' are equal to each other, and all the lower end bevels marked A 
and A' are equal;—the bevels and lengths of the braces being measured 
on the dotted lines extending from toe to toe; the slant heights of the 
intersections of the braces with the posts at A' and B' are found as 
already explained, by the use of the common rafter table, or by the use 
of the square. 








PLATE 4 1. 


This Plate represents an improved Straining-Beam Bridge, 50 feet 
long, for a Railroad. The principles of its construction are essentially 
the same as those of the bridges represented on Plate 30; the improve¬ 
ment of having the straining beams, stringers, and main braces, each in 
two pieces instead of one, bolted together, yet separated by packing blocks, 
as here shown, renders them much stronger and more durable; and the 
use of 3 inch planks, 10 inches wide, for counter braces, and the lateral 
braces, as shown in the floor-section, effectually meets the objections to 
this style of bridges, as stated on page 94; and they are probably as in¬ 
expensive and simple in construction as it is possible for a safe bridge 
to be built. 

Fig. 1 shows the side elevation, and Fig. 2, the floor-section, without 
ties or rails, to exhibit the manner of framing. The counter braces are 
let into each other, at their intersection, at the bottom, 1 inch each, and 
thus locked together; the supporting rods are crossed, as recommended 
under Plate 30. Bills of timber and iron, in detail, are subjoined. 

In the State of Illinois, the cost of a bridge of this kind would be 
about $10 per lineal foot. 

% 

Bill of Timber for One Span Improved Straining Beam Bridge, 50 feet long. 


2 Wall plates, 

10 by 12 

in., 

20 feet long, 

B. M. 

= 400 feet. 

8 Bolsters, 

10 by 10 

(( 

12 

(c 

(( 

a 

= 800 

u 

4 Stringers, 

7 by 14 

a 

50 

u 

u 

(( 

=1633 

a 

8 Main braces, 

7 by 10 

« 

16 

u 

a 

a 

= 747 

(( 

8 Minor “ 

7 by 8 

a 

14 

(( 

a 

« 

= 523 

(( 

6 Lateral u 

6 by 6 

u 

22 

(< 


a 

= 396 

a 

4 Straining beams, 

7 by 10 

<( 

24 

(( 

« 

u 

= 560 

(6 

25 Cross floor timbers, 

6 by 13 

« 

18 

(( 

(( 

<( 

=2925 

a 

4 Rail stringers, 

10 by 12 

a 

26 

(( 

a 

(( 

=1040 

u 

3 Needle beams, 

10 by 10 

a 

20 

(( 

(( 

a 

= 500 

u 

4 Counter braces, 

3 by 10 

« 

18 

(( 

(( 

u 

= 180 

u 

1 Packing, 

4 by 14 

(C 

12 

a 

(( 

(( 

= 56 

u 

i “ 

(116) 

3 by 10 

a 

12 

(( 

u 

u 

= 30 

9790 

a 

































































































































































































































' 






















. 


































BBIDGE BUILDING. 





117 



Bill of Iron. 






8 Supporting rods, 

if 

in. iron, 12 feet, 

4 

in. long, 

=812 

lbs. 

4 <( u 

n 

“ 12 « 

4 

u 


=210 

a 

4 Cross " 

1 

“ 17 “ 

6 

a 


=178 

u 

8 Brace bolts, 

f 

« 2 “ 

6 

(( 


= 30 

u 

18 “ “ 

i 

u 4 a 

8 

(( 


= 45 

u 

4 « a 


(( 4 a 

9 

u 


= 11 

u 

72 lbs. Nuts, 






= 72 

u 







1358 




Bill of Castings. 






16 Washers, 


6 in. diam. hole. 

i 

H 

in. 

= 96 

lbs. 

8 “ 


6 “ “ 


1 3 

8 

a 

= 48 

(( 

60 “ 


3 u “ 


8 


= 75 

a 

8 Lateral angle blocks, 





=160 

a 







379 





PLATE 42. 


Tli is Plate represents the manner of constructing and setting a circu¬ 
lar center of large dimension, for the purpose of springing a heavy stone 
arch. Let the outline of the frame be drawn upon a floor or platform, 
by taking OA as a radius, as described on page 67, Plate 15. It will 
be found convenient to cut a pattern of thin stuff for the 12 pieces of 
the circular rim, by which those pieces are to be made of 2 inch plank. 
One of them should be cut in two in the middle, and then all spiked 
together, lapped as represented in the plate. The 3 middle braces ex- 
ten I from the lower edge of the long brace to the upper edge of the rim, 
the middle braces to be let in the thickness of one plank, and at the 

lower end 1J inch on each side, and spiked to the rims and to the lower 

brace. The rims should not be more than 30 inches apart, and may be 
covered with good fencing laid close together. 

These large centers should be set to their places by wedges, as repre¬ 
sented in the plate, or they could not easily be taken out. 

Bill of Timber for One Rim of 28 feet span and 10 feet rise. 

12 pieces, 2 by 12 in., 7 ft. long. 

4 pieces, 2 by 10 in., 14 ft. long. 

3 braces, 6 by 6 in., 12 ft. long. 


( 118 ) 



























































































. 



















































































' 











































































































































































































































PLATE 43. 


This P .ate exhibits a plain and excellent plan for a frame-work to 
support 2 railroad water-tank. Fig. 1 represents the elevation as seen 
from the track, and Fig. 2 the foundation and bottom of the tank. 

The proper method of bracing between the posts is shown in Fig. 1, 
and the dotted lines on Fig. 2 show the proper places to brace. 

The workman should be careful to have the braces around the center 
all run the same way, the outer braces running up from right to left 
and, vice versa, the inner braces running up from left to right, passing 
each other at the center ; both series to be framed flush with the posts 
on their respective sides. The shoulders and bevels of all the braces 
must be framed the same way, so that changing the ends of a brace will 
not change its face side. The braces are let into the posts not less than 
1 inch to give support to the toe, and the posts, in this example, being 5 
feet 4 inches apart, the run of the braces will be 5 feet 6 inches^ and the 
length, on a regular run, as found in the brace table, page 141, is 7 feet 
9inches. Both the length and bevels of the braces are measured on 
the dotted lines extending from toe to toe in Fig. 1, the bevels being 
marked at 45° from these lines, for a regular run. These explanations 
are sufficient to enable the workman to construct such a husk or frame 
without further instruction. Bills of timber and iron are given in 
detail. 

Height and Location of Tank. 

The height from the top of the rail on the track to the top of the 
floor timbers on the frame work, when the water is to be taken out of 
the bottom of the tank, in the usual way, by the use of a goose-neck 
valve, is 11 feet; but when the water is’ to be taken from the side, it 
should be about 1 foot lower. 

For a tank of 22 feet outside diameter, the distance of the center of 
the tank from the middle of the track should be 21 feet. 

To Estimate the Capacity of a Tank. 

Carpenters are frequently asked to give the capacity of a tank or 
cistern, which it is difficult for them to do without some rule or data, 
and we will therefore give the capacity of this tank and the rule for 

(1191 


120 


CARPENTRY MADE EASY. 


obtaining it. The inside dimensions, after deducting the thickness of 
the staves, bottom and chime, are as follows: 

Top diameter, 20 ft. 8 in. 

Bottom diameter, 21 ft. 6 in. 

Mean diameter, 21 ft. 2 in. 

Height, 15 ft. 4 in. 

To obtain the mean diameter add § of the difference between the top 
and bottom diameters to the top diameter; and to find the capacity, multi¬ 
ply the square of the mean diameter, in inches, by the height, in inches, 
and this product by .0034: the answer will be in gallons. This decimal 
.0034 is obtained by dividing the decimal .7854 by 231, the number of 
cubic inches in a gallon; and the decimal .7854 is used to reduce the 
capacity of a square tank to that of a circular one, for .7854 is the area 
of a circle whose diameter is one. 

To make it more plain we give the figures in full— 

Mean diameter = 254 in. 

Height =184 in. 

254 x 254 x 184 x .0034 = 40361.20 gallons. 

Operation: 254 

254 

1016 

1270 

508 

64516 

184 

258064 

516128 

64516 

11870944 

.0034 

47483776 

35612832 

40361.2096 Ans. 

To reduce gallons to barrels divide by 31J. 

To find the circumference of a tank, if we know the diameter, we 
multiply the diameter by 3.1416, or if we know the circumference we 
can find the diameter by dividing it by 3.1416; because this is the cir- 








RAILROAD WATER-TANK. 


121 


cumference of a circle whose diameter is 1. Suppose we want the length 
of the bottom hoop on a tank 22 feet across the bottom— 

Operation: 3.1416 

_22 

62832 

62832 

69.1152 Am. 69 ft. 1.38 in. 

_12 

1.3824 

To find the capacity of a square cistern, multiply its three dimensions 
together, viz., its length, breadth and height in inches, and divide the 
product by 231, and the answer will be in gallons. 


Bill of Timber for Plate 43. 



2 Caps, 12 by 12 

in. 

24 feet long, Board 

measure 

= 576 

2 Caps, 12 by 12 

« 

18 

« 

<( 

u 

u 

= 432 

12 Posts, 12 by 12 

u 

12 

« 

u 

a 

u 

=1728 

6 Floor timbers 6 by 12 

u 

24 

a 

(( 

u 

(( 

= 864 

2 “ “6 by 12 

a 

22 

u 

u 

u 

(( 

= 264 

2 “ “6 by 12 

u 

20 

a 

a 

(( 

(( 

= 240 

2 “ “ 6 by 12 

u 

18 

u 

a 

u 

(( 

= 216 

2 “ “6 by 12 

a 

16 

a 

« 

(( 

u 

= 192 

2 “ “6 by 12 

u 

12 

a 

(( 

u 

(( 

= 144 

16 Braces, 6 by 6 

u 

8 

a 

<( 

(( 

u 

= 384 

20 Bib pieces, 3 by 6 

(( 

16 

u 

<( 

a 

a 

= 480 








5520 

Bill of Iron 

for 

Same. 



16 Bods, 7 ft. 8 in. long, inch 

iron, 


weight, 

324 lbs. 

32 Cast Washers, 2 lbs. each, 





64 lbs. 

32 Nuts, 1 lb. each, 







32 lbs. 








420 lbs. 


u 

u 







PLATE 44. 


This Plate illustrates the proper manner of framing and constructing 
pile bridges. This is a difficult and perplexing job to inexperienced 
workmen, because the piles cannot be driven exactly in line nor at ex¬ 
actly equal distances apart, and the frame must, therefore, be fitted to 
them. After the piles are driven, we first saw them off to the proper 
Height for the top of the tenons, and then level a bat-board to the height 
of the shoulder and nail it to one of the outside piles, as represented in 
Fig. 1; then level the other outside pile, in the same manner, and lay a 
straight edge upon them on each side of the bent; and nail bat-boards, 
by it, upon both sides of each pile. Then strike a chalk line on the 
tops of the piles, and lay out our tenons; then place the square upon the 
bat-boards and mark the sides of the tenons. 

Fig. 2 shows one bent of piles cut to the top of the tenons and ready 
for sawing the shoulders. To do this, place the saw flat upon the bat- 
boards, saw the shoulders, and dress off the tenons, sizing their edges 
3 inches thick and as wide as they will square, then lay a board upon 
the shoulders of the piles, along the whole width of the bent, and mark 
carefully at each edge of the tenons; this board applied to the cap will 
give the proper places for the tenons, their lengths and their distances 
apart. The draw-bores should be 1J inches in diameter, and the centers 
24 inches from the edge of the tenon and 1|- inches from the shoulder. 
After the cap has been faced to the proper size, mark the draw-bores 
upon it 2J inches from the ends of the mortices and 2 inches from the 
face, which gives a draw of J inch. 

By framing each bent in this manner, the caps will rest square upon 
the shoulders of the piles and make a perfect fit, securing both greater 
strength and greater durability than can be done in any other manner; 
for it is not uncommon to see these bridges framed so loosely that the 
whole hand can be inserted between the shoulder and the cap. 

Fig. 3 represents one bent of a pile bridge finished; with lateral 
braces bolted to the piles to prevent side vibration and to give greater 
stability to the work. 

Fig. 4 shows the side elevation, the ends of the ties and the manner 
of framing the guard-rail. The ties are 5 by 8 inches, except every 4th 
tie, which should be 6 by 8 inches, and let down over the stringer with 
( 122 ) 










































































































































































' 



























































BRIDGE BUILDING. 


123 


a shoulder of 1 inch on each side, to prevent the track from slipping on 
the bridge, and on curves the stringers should be secured to the caps by 
long spikes. 

The guard-rail should be 5 by 6 inches and cut out 2 inches over each 
tie, and fastened by wrought-iron spikes driven through guard-rail and 
tie into the stringer. The guard-rail is used for the double purpose of 
strengthening the bridge and preventing the train from getting off the 
bridge should it jump the track. The ties are placed 6 inches apart, or 
14 inches from center to center. 

Fig. 5 represents the floor section without either ties or guard-rails, 
in order to show the manner of securing the ends of the rail-stringers. 
This is done by inserting a 3 inch oak plank about 3 feet long, as rep¬ 
resented in the plate, secured by f inch bolts 9 inches from the ends of 
the stringers, and 3 inches from the top and the bottom, on each side of 
the joint, making 8 bolts to each bent. 








EXPLANATION OF THE TABLES. 


Definitions of Terms and Phrases used in this Explanation, and in other 

Places in this Work. 

The length of a rafter is understood to be measured from the ex¬ 
treme point of the foot to the extreme point of its upper end * But 
in these Tables no allowances are made for the projection of rafters be¬ 
yond the plate, or for ridge poles; so that the length of common rafters 
is understood to be the distance from the upper and outer corner of the 'plate 
to the very peak of the roof 

The RUN of a rafter is the horizontal distance from the extreme 
point of the foot to a perpendicular let fail from the upper end. In 
common roofs } the run of the rafters is half the width of the building. 

The RISE of a rafter is the perpendicular distance from the upper 
end of the rafter to the level of the foot. 

The gain of a rafter is the difference between its run and its length. 
For example, a rafter whose run is 12 feet, and whose length is 13 
feet, has 1 foot gain. 

The learner will easily perceive that the length of any rafter is the 
hypotenuse of a right-angled triangle, of which its run and its rise are 
the other two sides. The length is therefore ascertained with perfect 
accuracy by adding the square of the run to the square of the rise, 
and extracting the square root of their sum. (See Part I., Prop. 
XXIV.) 

Example 1. The length of a common rafter is required in a build¬ 
ing 24 feet wide, the roof of which is desired to have a pitch of 5 
inches to the foot. The run is therefore 12 feet, the square of which 


* Except in hip rafters , the length of which is always to be measured on the backing , 
or along the middle line of the upper surface; for when the side bevel is all cut on one 
side of the upper end, as it sometimes is, then the point of the rafter will extend half its 
thickness beyond its estimated length, as given in the table, &c. 


( 127 ) 






128 


CARPENTRY MADE EASY. 


is the product of 12 multiplied by 12, or 144 feet. The rise is 12 

times 5 inches, or 60 inches, or 5 feet, the square of which is 25 

feet; which, added to 144, makes 169 feet, of which we extract the 

square root thus: . . 

(the rule for which may be found in any common- 1)169(13 

school arithmetic) and find it to be 13 feet, the exact ^_ 

length of the rafter required. 23)69 

^ 69 

00 

But in most cases the result is obtained in the form of a fraction; 
and it will be found convenient to reduce the run and the rise to inches, 
in the first place, and then the root is obtained in inches and decimals 
of an inch, which can be carried out to any degree of accuracy re¬ 
quired. T.n these Tables they are carried to hundredths of an inch. 


Example 2. Required the length of a rafter for the building de¬ 
scribed in Plate 4 of this work. Width of building, 12 feet; rise of 
rafter, 6 inches to the foot. 

The run of the rafter is 6 feet, or 72 in., of wbicb the square is 5184 

1296 

6480 


rise 


u 


u 


or 36 


u 


u 


u 


and their sum is 

of which we proceed to extract the square root thus: 
and find it to be 80 inches and 49 hundredths of an 
inch; or, 6 feet 8 inches and , 4 0 9 G , as given in the 
Table, which is the exact length of the required rafter. 


8)6480(80.49 

64 


1604)8000 

6416 


16089)158400 

144801 


13599 







TABLE I. 


The t se <£ this Table is to furnish the practical carpenter with the 
precise lengths of common rafters for buildings of all sizes, and for 
roofs of every pitch. The Table is carried out to buildings of 60 feet 
in width; but should the length of rafters for a wider building be 
required, it will be necessary to add such numbers together in the left- 
hand column as will make their sum equal to the width of the build¬ 
ing, and then the sum of the lengths of the rafters given in the Table 
opposite these numbers, thus added together, will be the true length 
of the rafters required. 

For example, suppose it were required to find the length of rafters 
for a building 84 feet wide, 6-inches rise. We find the length of a 
rafter of half that width, of the same pitch, to be 23 ft. 5.74 in., and 
double this number would be the length of the rafter required, or 
46 ft. 11.48 in. 

Example 2. Required the length of the rafters for a building 102 
feet wide, 5 inches rise. We perceive that 50 and 52 added together 
will make 102. The lengths of these two dimensions for this pitch, 
as given in the Table, are 27 ft. 1 in., and 28 ft. 2 in., the sura of 
which is 55 ft. 3 in., the length of the rafters required. 

9 


( 129 ) 


130 


CARPENTRY MADE EASY, 


TABLE I. 

Length of Rafters m Feet, Inches, and Hundredths of an Inch. Rise of Rafter 

to the Foot in Inches. 


Width of 
Building 
in Feet. 

c 

§ S- 

M ° « 

W M 

1 inch 
Rise. 

2 inch 
Rise. 

3 inch 
Rise. 

A inch 
Rise. 

6 inch 
Rise. 

6 inch 
Rise. 

7 inch 
Rise. 

8 inch 
Rise. 

9 'nch 

R ise. 

4 

2 

2 : 0.0S 

2 : 0.33 

2: 0.73 

2: 1.29 

2: 2.00 

2 : 

2.83 

2: 3.78 

2: 4.84 

2 : 

6 

5 

2:6 

2: 6.10 

2 : 6.41 

2: 6.92 

2: 7.62 

2: 8.50 

2: 

9.54 

2 : 10.73 

3: 0.05 

3 : 

1.50 

6 

3 1 

i 

3 : 0.12 j 

1 

3 : 0.49 

3: 1.10 

3: 1.94 

3: 3 

3 : 

4.24 

3: 5.67 

3 : 7.26 

3 : 

9 

7 1 

3 : 6 

3: 6.14 

3 : 6.57 

3 : 7.30 

3: 8.27 

3: 9.50 

3 : 

10.95 

4: 0.62 

4: 2.47 

4 : 

4.50 

8 

4 

4: 0.16 

4 : 0.66 

4: 1.47 

4: 2.59 

4: 4 

4 : 

5.66 

4: 7.56 

4: 9.68 

5 : 

9 

4 : 6 

4: 6.18 

4 : 6.74 

4: 7.66 

4: 8.92 

4 : 10.50 

5 : 

0.36 

5: 2.51 

5: 4.S9 

5 : 

7.50 

10 | 

5 1 

5 : 0.20 

5 : 0.82 

5: 1.84 

5 : 3.24 

5 : 5 

5 : 

7.OS 

5: 9.45 

6 : 0.10 

6 : 

3 

11 

5 : 6 | 5 : 6.22 

5 : 6.91 

5: 8.03 

5: 9.57 

5 : 11.50 

6 : 

1.78 

6: 4.41 

6: 7.31 

6 : 

10.50 

12 

6 

6 : 0.25 

6 : 0.99 

6: 2.21 

6: 3.89 

6: 6 

6 : 

8.49 

6 : 11.34 

7 : 2.52 

7 : 

6 

11 

7 

7 : 0.29 

7 : 1.15 

7 : 2.58 

7: 4.54 

7: 7 

7 : 

9.91 

8: 1.23 

8 : 5.94 

8: 

9 

16 

8 

S : 0.33 

S : 1.32 

8: 2.95 

8: 5.19 

8: 8 

8 : 

11.32 

9 : 3.12 

9 : 7.36 

10 : 

18 

9 

9 : 0.36 

9 : 1.48 

9: 3.32 

9 : 5.84 

9 : 9 

10 : 

0.74 

10: 5.03 

10: 9.79 

11: 

3 

20 

10 

10 : 0.40 

1 

10 : 1.61 

10 : 3.69 

10 : 6.49 

10 : 10 

11 : 

2.16 

11 : 6.92 

12: 0.22 

12: 

6 

22 

H 

jll : 0.14 

11 : 1.81 

11 : 4.06 

11 : 7.14 

11 : 11 

12 : 

3.58 

12: 8.81 

13: 2.64 

13 : 

9 

24 

12 

,12 : 0.49 

12 : 1.98 

12: 4.43 

12: 7.79 

13: 

13 : 

4.99 

13 : 10.70 

14: 5.06 

15: 

26 

13 

13 : 0.53 

13 : 2.15 

13: 4.80 

13: 8.44 

14: 1 

14 : 

6.41 

15 : 0.59 

15: 7.4S 

16: 

3 

28 

14 

14 : 0.57 

14 : 2.31 

14: 5.17 

14: 9.09 

15: 2 

15 : 

7.82 

16: 2.48 16: 9.90 

17 : 

6 

30 

15 

15 : 0.62 115 : 2.48 

15: 5.54 

15: 9.73 

16: 3 

16 : 

9.24 

17: 4.38 

IS: 0.32 

18: 

9 

32 

16 

16 : 0.66 

16 : 2.64 

16: 5.91 

16 : 10.38 

17: 4 

17 : 

10.65 

18: 6.27 

19: 2.74 

20 : 

31 

17 

17 : 0.70 

17 : 2.80 

17 : 6.28 

17 : 11.03 

18: 5 

19 : 

0.07 

19 ; 8.16 

20: 5.16 

21: 

3 

36 

18 

18 : 0.73 

18 : 2.96 

IS: 6.65 

18 : 11.68 

19: 6 

20 : 

1.48 

20 : 10.06 

21 : 7.58 

22: 

6 

38 

19 

19 : 0 78 

19 : 3.12 

19 : 7.02 

20 : 0.33 

20: 7 

21 : 

2.90 

21 : 11.95 

22: 10 

23 : 

9 

40 

20 

j20 : 0.81 

20 : 3.28 

20: 7.39 

21 : 0.98 

21 : 8 

22: 

4.32 

23: 1.85 

24 : 0.43 

25 : 

42 

21 

21 : 0.85 

21 : 3.45 

21 : 7.76 

22: 1.63 

22: 9 

23: 

5.74 

24: 3.74 

25 : 2.85 

26: 

3 

44 

22 

22 : 0.89 

22 : 3.62 

,22: 8.12 

23: 2.2S 

23 : 10 

24 : 

7.16 

25: 5.64 

26: 5.27 

27 : 

6 

46 

23 

23 : 0.93 

23 : 3.79 

23: 8.49 

21: 2.93 

24 : 11 

25 : 

8.58 

26 : 7.53 

27 : 7.69 

28: 

9 

48 

1 24 

21 : 0.97 

i 

24 : 3.96 

24: 8.86 

25 : 3.57 

26: 

26 : 

10 

27 : 9.42 

28 : 10.12 

30 : 

50 

25 

25 : 1.01 

25 : 4.13 

25: 9.23 

26: 4.22 

27 : 1 

27 : 

11.41 

,28 : 11.31 

30 : 0.54 

31 : 

3 

52 

26 

26 : 1.06 

1 

j26 : 4.30 

26: 9.60 

27: 4.87 

28: 2 

29 : 

0.83 30: 1.20 

31 : 2.96 

32 : 

6 

54 

27 

27 : 1.10 

i 

27 : 4.46 

27 : 9.97 

28: 5.52 

29: 3 

30 : 

2.24 

31 : 3.09 

32; 5.39 

33 : 

9 

66 

28 

28: 1.14 

28 : 4.63 

28 : 10.34 

j 29 : 6.17 

30: 4 

31 : 

3.66 

32: 4.98 

33: 7.81 

35: 

58 

29 

29 : 1.18 

29 : 4.79 

29 : 10.71 

30: 6.S2 

31 : 5 

32: 

5.08 

33: 6.87 

34 : 10.24 

36: 

3 

60 

| 30 

30 : 1.23 

30 : 4.96 

30 : 11.08 

31 : 7.47 

32: 6 

33 : 

6.49 

34: 8.76 

36; 0.66 

.37 : 

6 








































































































































































































































































































































































































































































































EXPLANATION OF TABLES 


131 


TABLE I.— Continued. 

Length of Eafters in Feet, Inches, and Hundredths of an Inch. R. se of Rafter 

to the Foot in Inches. 


Width of 
Building 
in Feet. 

Bun 

of 

Bafter. 

. _ 1 

10 inch 
Bise. 

11 inch 
Bise. 

12 inch 
Bise. 

13 inch 
Bise. 

14 inch 
Bise. 

IS inch 
Bise. 

16 inch 
Bise. 

17 inch 
Bise. 

18 inch 
Bise. 

4 

2 

2 : 

7.24 

2 : 

8.55 

2 : 

9.94 

2 : 11.38 

2 : 

0.87 

3 : 2.41 

3 : 

4 

3 : 5.61 

3 : 7.26 

5 

2 : 6 

1 

3 : 

3.05 

3 : 

4.69 

3 : 

6.42 

3 - 8.22 

3 : 

10.09 

4 : 0.02 

4 : 

2 

4 : 4.02 

4 : 6.08 

e 

3 

3 : 

10 . S 6 

4 : 

0.83 

4 : 

2.91 

4 : 5.07 

4 : 

7.31 

4 : 9.62 

5 : 

5 : 2.42 

5 : 4.89 

7 

3 : 6 

i 

4 : 

6.67 

4 : 

8.97 

4 : 

11.39 

5 : 1.92 

5 : 

4.53 

5 : 7.23 

5 : 

10 

6 : 0.82 

6 : 3.71 

8 

4 

5 : 

2.48 

5 : 

5.11 

5 : 

7.88 

5 : 10.76 

6 : 

1.75 

6 : 4.83 

6 : 

8 

6 : 11.23 

7 : 2.53 

9 

4 : 6 

5 : 

10.29 

6 : 

1.25 

6 : 

4.36 

6 : 7.61 

6 : 

10.97 

7 : 2.44 

7 : 

6 

7 : 9.63 

8 : 1.34 

10 

5 

6 : 

6.10 

6 : 

9.39 

7 : 

0.85 

7 : 4.45 

7 : 

8.19 

8 : 0.04 

8 : 

4 

8 : 8.04 

9 : 0.16 

11 

5:6 

7 : 

1.91 

7 : 

5.53 

7 : 

9.33 

8 : 1.30 

8 : 

5.41 

8 : 9.65 

9 : 

2 

9 : 6.44 

9 : 10.98 

12 

6 

7 : 

9.72 

8 : 

1.67 

8 : 

5.82 

8 : 10.15 

9 : 

2.63 

9 : 7.25 

10 : 

10 : 4.84 

10 : 9.79 

14 

7 

9 : 

1.34 

9 : 

5.95 

9 : 

10.79 

10 : 3 . S 4 

10 : 

9.07 

11 : 2.46 

11 : 

8 

12 : 1.65 

12 : 7.43 

16 

8 

10 : 

4.96 

10 : 

10.23 

11 : 

3.76 

11 : 9.53 

12 : 

3.51 

12 : 9.67 

13 : 

4 

13 : 10.46 

14 : 5.06 

18 

9 

11 : 

8.58 

12 : 

2.51 

12 : 

8.73 

13 : 3.22 

13 : 

9.95 

14 : 4.88 

15 : 

15 : 7.27 

16 : 2.69 

20 

10 

IS : 

0.20 

13 : 

6.79 

14 : 

1.70 

14 : 8.91 

15 : 

4.39 

16 : 0.09 

16 : 

8 

17 : 4.08 

18 : 0.33 

22 

11 

14 : 

3.82 

14 : 

11.07 

15 : 

6.67 

]6 : 2.60 

16 : 

10.83 

17 : 7.30 

18 : 

4 

19 : 0 . 8 S 19 : 9.96 

24 

12 

15 : 

7.44 

16 : 

3.34 

16 : 

11.64 

17 : 8.30 

18 : 

5.27 

19 : 2.51 

20 : 

20 : 9.69 

21 : 7.59 

26 

13 

16 : 

11.06 

17 : 

7.62 

18 : 

4.61 

19 : 1.99 

19 : 

11.71 

20 : 9.72 

21 : 

8 

22 : 6.50 j 23 : 5.22 

28 

14 

18 : 

2.68 

18 : 

11.90 

19 : 

9.58 

20 : 7.67 

21 : 

6.14 

22 : 4.93 

23 : 

4 

24 : 3.31 

25 : 2.86 

30 

15 

19 : 

6.30 

20 : 

4.18 

21 : 

2.55 

22 : 1.37 

23 : 

0.58 

24 : 0.14 

25 : 

26 : 0.12 

27 : 0.49 

32 

16 

20 : 

9.92 

21 : 

8.46 

22 : 

7.52 

23 : 7.07 

24 : 

7.02 

25 : 7.35 

26 : 

8 27 : 8.92 

| 

2 S : 10.12 

31 

17 

22 : 

1.54 

23 : 

0.74 

24 : 

0.49 

25 : 0.77 

26 : 

1.46 

27 : 2.56 

28 : 

4 

29 : 5.73 

30 : 7.76 

36 

18 

23 : 

i - . 

5.16 

24 : 

5.02 

25 : 

5.47 

26 : 6.46 

27 : 

7.90 

28 : 9.77 

30 : 

31 : 2.55 

32 : 5.39 

38 

19 

' 24 : 

1 

8.78 

25 : 

9.30 

26 : 

10.44 

28 : 0.15 

29 : 

2.34 

30 : 4.98 

31 : 

8 

32 : 11.36 

34 : 3.02 

40 

20 

26 : 

0.40 

27 : 

1.57 

28 : 

3.41 

29 : 5.84 130 : 

8:70 

32 : 0.19 

33 : 

4 

34 : 8.17 

36 : 0.66 

42 

21 

27 : 

4.02 

28 : 

5.85 

29 : 

8.38 

30 : 11.53 

32 : 

3.22 

33 : 7.39 

35 : 

36 : 4.97 

37 : 10.29 

44 

22 

2 S : 

7.64 

29 : 

10.13 

31 : 

1.35 

32 : 5.22 

33 : 

9.66 

35 : 2.60 

36 : 

8 

38 : 1.78 

39 : 7.92 

46 

23 

29 : 

11.26 

31 : 

2.41 

32 : 

6.32 

33 : 10.92 

35 : 

4.10 

36 : 9.81 

38 : 

4 39 : 10.59 

41 : 5.55 

48 

24 

31 : 

2.88 

32 : 

6.69 

33 : 

11.29 

35 : 4.61 

36 : 

10.54 

38 : 5.02 

40 : 

41 : 7.40 

43 : 3.19 

50 

25 

32 : 

6.50 

33 : 

10.97 

35 : 

4.26 

36 : 10.30 

38 : 

4.98 

40 : 0.23 

41 : 

8 

43 : 4.21 

45 : 0.82 

52 

26 

33 : 

10.12 

35 : 

3.25 

36 : 

9.23 

38 : 3.99 

39 : 

11.42 

41 : 7.44 

43 : 

4 

45 : 1.01 

46 : 10.45 

54 

27 

( 3 o i 

1.74 

36 : 

7.53 

38 : 

2.20 

39 : 9.68 

41 : 

5.86 

43 : 2.65 

45 : 

46 : 9.82 

48 : 8.09 

56 

28 

36 : 

i 

5.36 

37 : 

11.80 

39 : 

7.17 

41 : 3.37 

43 : 

0.30 

44 : 9 . S 6 

46 : 

8 

48 : 6.63 

50 : 5.72 

58 

29 

37 : 

8.98 

39 : 

4.08 

41 : 

0.14 

42 : 9.06 

44 : 

6.74 

46 : 5.07 

48 : 

4 50 : 3.44 

52 : 3.35 

60 

30 | 

39 : 

0.60 

40 : 

8.36 

42 : 

5.11 

44 : 2.75 

46 : 

1.18 

48 : 0.28 

50 : 

52 : 0 . 2 » 

54 : 9.19 
































































































































































































































































































































































































































































TABLE II. 


♦ 


Length of Hip Rafters. 

# 

If a roof were perfectly horizontal or flat, the hip rafters would each 
be equal to the diagonal of a square, haying for its side half the width 
of the building; and the square root of twice the square of half the 
side, would, in that case, be the length of the hip rafter. This we call 
the RUN of the hip rafter. But if the roof has any pitch, the length of 
the rafter is greater than its run , and is always equal to the hypotenuse 
of a right-angled triangle, having the run for a base, and the rise for a 
perpendicular; and the length is found, as in common rafters, by add¬ 
ing the square of the run to the square of the rise, and extracting the 
square root of the sum. 

Two calculations are necessary according to the above demonstra¬ 
tion: First, for obtaining the run; and secondly, having found the run, 
from that to obtain the length. 

First. Suppose the width of the building to be 40 feet, and the rise 
of the roof 5 inches to the foot, or 100 inches; then half the width of 
the building, 20 feet, is 240 inches, the square of which is 57,600. 
Double this number (for the two sides of the square) is 115,200 inches, 
of which the square root is 339 ^ inches, or, 28 ft. 3 y 4 0 V in., which 
is the run of the hip rafter. 

Second. To obtain the length , which equals the hypotenuse of a 
right-angled triangle, of which the run is the base and the rise the 
perpendicular. 

The run is 339.41 inches as obtained above, 

The square of which is 115,200 

The rise is 100 inches, of which the square is 10,000 

The sum of these two squares is 125,200 inches, of which the square 
root is 353.83 inches, or 29 ft. 5.83 in., which is the true length of the. 
hip rafter. 

The process of obtaining the length is explained above according to 
the long ivay, and the most obvious and analytical way also, and one 
which every practical mechanic should make himself fully familiar 
with : but practically , the process may be shortened as follows:— 
(132) 


I 


EXPLANATION OF TABLES. 


133 


Add the square of the rise to twice the square of half the width ) and the 
square root of the sum will be length of hip rafter required. Thus:— 

The square of the rise (100 inches) is 10,000 

Twice the square of half the width is 115,200 

Their sum is 125,200 

of which the square root is 353.83 in., or 29 ft. 5.83 in. as before; 
which is the true length of the hip rafter as measured on the backing. 
See note on p. 127. 


TABLE II. 


Length of Hip Rafters in Feet, Inches, and Hundredths of an Inch. 


■ ■■■— 

Width of 
Building 
in feet. 

a 

3 

« 

of 

Rafter. 

1 inch 
Rise. 

2 inch 
Rise. 

S inch 
Rise. 

4 

inch 

Rise. 

6 inch 
Rise. 

6 inch 
Rise. 

1 inch 
Rise. 

• 

8 inch 
Rise. 

4 

2 : 

9.94 

2 : 10 

2 : 10.17 

2 : 10.46 

2 

: 10.87 

2 : 11.38 

3 : 

3 : 

0.71 

3 : 

1.52 

5 

3 : 

6.42 

3 : 6.50 

3 : 6.72 

3 : 7.08 

3 

: 7.58 

3 : 8.22 

3 : 9 

3 : 

9.89 

3 : 

10.90 

6 

4 : 

2.91 

4 : 3 

4 : 3.26 

4 :. 3.70 

4 

: 4.30 

4 : 5.07 

4 : 6 

4 : 

7.07 

4 : 

8.28 

7 

4 : 

11.39 

4 : 11.50 

4 : 11.80 

5 : 0.31 

5 

: 1.02 

5 : 1.92 

5 : 3 

5 : 

4.25 

5 : 

5.66 

8 

5 : 

7.88 

5 : 8 

5 : 8.35 

5 : 8.93 

5 

: 9.74 

5 : 10.76 

6 : 

6 : 

1.43 

6 : 

3.04 

9 

6 : 

4.36 

6 : 4.50 

6 : 4.89 

6 : 5.55 

6 

: 6.46 

6 : 7.61 

6 : 9 

6 : 

10.60 

7 : 

0.42 

10 

7 : 

0.85 

7 : 1 

7 : 1.44 

7 : 2.16 

7 

: 3.17 

7 : 4.45 

7 : 6 

7 : 

7.78 

7 : 

9.80 

11 

7 : 

9.33 

7 : 9.50 

7 : 9.98 

7 : 10.78 

7 

: 11.89 

8 : 1.30 

8 : 3 

8 : 

4.96 

8 : 

7.18 

12 

8 : 

5.82 

8 : 6 

8 : 6.52 

8 : 7.39 

8 

: 8.61 

8 : 10.15 

9 : 

9 : 

2.14 

9 : 

4.56 

14 

9 : 

10.79 

9 : 11 

9 : 11.61 

10 : 0.63 

10 

: 2.04 

10 : 3.84 

10 : 6 

10 : 

8.50 

10 : 

11.33 

16 

11 : 

3.76 

11 : 4 

11 : 4.70 

11 : 5.86 

11 

: 7.48 

11 : 9.53 

12 : 

12 : 

2.86 

12 : 

6.09 

18 

12 : 

8.73 

12 : 9 

12 : 9.79 

12 : 11.09 

13 

: 0.91 

13 : 3.22 

13 : 6 

13 : 

9.21 

14 : 

0.85 

20 

14 : 

1.70 

14 : 2 

14 : 2.88 

14 : 4.33 

14 

: 6.35 

14 : 8.91 

15 : 

15 : 

3.57 

15 : 

7.61 

22 

15 : 

6.67 

15 : 7 

15 : 7.96 

15 : 9.57 

15 

: 11.79 

16 : 2.60 

16 : 6 

16 : 

9.93 

17 : 

2.37 

24 

16 : 

11.64 

17 : 

17 : 1.05 

17 : 2.80 

17 

: 5.22 

17 : 8.30 

18 : 

18 : 

4.29 

18 : 

9.34 

26 

18 : 

4.61 

18 : 5 

18 : 6.14 

18 : 8.03 

18 

: 10.66 

19 : 1.99 

19 : 6 

19 : 

10.64 

20 : 

3.90 

28 

19 : 

9.58 

19 : 10 

19 : 11.23 

20 : 1.26 

20 

: 4.09 

20 : 7.68 

21 : 

21 : 

5 

21 : 

10.66 

30 

21 : 

2.55 

21 : 3 

21 : 4.32 

21 : 6.50 

21 

: 9.53 

22 : 1.37 

22 : 6 

22 : 

11.36 

23 : 

5.42 

32 

22 : 

7.52 

22 : 8 

22 : 9.40 

22 : 11.73 

23 

: 2.96 

23 : 7.06 

24 : 

24 : 

5.72 

24 : 

11.18 

34 

24 : 

0.49 

24 : 1 

24 : 2.49 

24 : 4.97 

24 

: 8.40 

25 : 0.76 

25 : 6 

26 : 

0.07 

26 : 

5.94 

36 

25 : 

5.46 

25 : 6 

25 : 7.58 

25 : 10.20 

26 

: 1.83 

26 : 6.45 

27 : 

27 : 

6.43 

28 : 

0.70 

38 

26 : 

10.44 

26 : 11 

27 : 0.67 

27 : 3.43 

27 

: 7.27 

28 : 0.14 

28 : 6 

29 : 

0.79 

29 : 

7.47 

40 

28 : 

3 . 4 '. 

28 : 4 

28 : 5.76 

28 : 8.27 

29 

: 0.71 

29 : 5.83 

30 : 

30 : 

7.15 

31 : 

2.23 



































































































































































































































































TABLE III. 


Hip and Jack Rafters on Octagonal Roofs. 

The length of one side of an octagonal building being commonly 
gi^en as the basis of calculation in framing, it will first be necessary 
from this basis, to determine with accuracy the width of the building 
from the middle of one side to the middle of the opposite side; and 
also the diagonal width, from one corner to the opposite corner. 

The width FG (in Plate 20, Fig. 1) is obviously the same as one 
side of the circumscribed square DE; and DE is made up of three 
parts, namely, DA, AB, and BE, one of which parts, AB, is known— 
being a side of the given octagon. The other two parts are equal to 
each other, namely, DA=BE * We have, therefore, to find the length 
of DA, to double it, and to add AB to it in order to ascertain the 
width of the building. The length of DA is found as follows:— 

In the right-angled triangle CAD, the hypotenuse AC, being one 
of the sides of the given octagon, is known; and the square of this 
hypotenuse is equal to the sum of the squares of the other two sides 
DA and DC, or to double the square of DA. 

For example, suppose the sides of the regular octagon be given 
equal to 16 feet, which, on reducing it to inches to insure greater accu¬ 
racy, is 192 inches. 

The square of 192 inches is 36,864 inches, one half which is 18,432 
inches, which is the square of DA. 

The square root of 18,432 inches is 135.76 inches, or 11 ft. 3.76 in., 
the length of DA. 

Double this number (for DAxBE), and add 16 feet for the length 
of AB, and we have 38 ft. 7.52 in., the width of the building. 

The diagonal width is obtained as follows: 

Let O' represent the point at the foot of the perpendicular let fall 


• The equality of DA and BE maybe demonstrated thus:—Suppose the figure di¬ 
vided into two parts by the line FG, and these two parts to be folded together, the line 
FG forming the fold; then the point A would fall upon the point B, the point C upon the 
point H, and the point D upon the point E; otherwise, the polygon is not a regular poly¬ 
gon, nor the circumscribed square a perfect square. In a similar manner, it may be 
demonstrated that the line AD is equal to DC, by supposing the figure to be folded upon 
the line DL. 

(134) 




EXPLANATION OF TABLES. 


135 


from 0, the apex of the roof, upon the plane or level of the plates C A, 

AB, &c. 

Then, in the right-angled triangle O'FA, the sum of the squares of 
the two sides AF and FO' will equal the square of the hypotenuse 
AO' 

Having found above that FG=38 ft. 7.52 in., then FO' will 
equal half this number, or 19 ft. 3.76 in., or 231.76 inches, the 
square of which is 53,712.6976 inches. 

FA is half of the given side AB, and is 8 feet, or 96 inches, of which 
the square is 9,216 inches, which, being added to 53,712.6976 inches, 
is 62,928.6976 inches, the square root of which is 250.85 inches, or 
20 ft. 10.85 in., the length of AO', or half the diagonal width of the 
building. Double this number is 41 ft. 9.70 in., the length of AP, 
the diagonal width required. 

Half the diagonal width is of course the run of the hip rafters , and half 
the square width is the run of the middle jack rafters; and, having ascer¬ 
tained these, the lengths of the rafters are calculated according to the 
rule given at the commencement of this general explanation of the 
Tables—by taking the square root of the sum of the squares of the run 
and the rise of any given rafter. 


E 


II. Width of Octagon given to find the Diagonal and the Side. 

It sometimes happens, as in church spires for example, that the 
width of an octagon is given, from which the other dimensions must 
be found. 

Let PC, the width of a regular octagon, 
be given, to find AB the side, and AE 
the diagonal. 

Draw OD from the centre of the octa¬ 
gon to an angle of the circumscribed 
square. Then OD 2 =CD 2 +OC 2 , or 20C 2 , 
since OD is the hypotenuse of the tri¬ 
angle ODC, of which the other two sides 
OC and DC are equal to each other, and 
each one equals half the given width of 
the octagon. Then, since O A bisects the vertical angle of the triangle 
COD, it divides the base into two segments, which are proportional to 
the adjacent sides (Part I., Prop. XXVI.); and we have the following 
proportion: 

DO : OC :: DA : AC; 









136 


CARPENTRY MADE EASY. 


and, by composition, 

DO+OC : OC :: DA+AC : AC; 
but AC is half tlie required side AB. 

Having obtained, by the above formula, the length of AC, it will 
be easy to obtain that of OA, since OA 2 =OC 2 +AC 2 . 

f 

.Example. Suppose PC, the given width, equals 10 ft. 8 in., which 
is the width of the base of the church spire described in Plate 
28. Reducing this number to inches, to insure greater accuracy, we 
have PC=128 inches. And OD would then equal the square root of 
the sum of the squares of OC and DC, eacli of which equals 64 inches. 
Double the square of 64 inches equals 8192 inches, the square root of 
which is 90.51 inches, which is the length of OD; then, by applying 
or substituting this value in the first proportion given above, we have 

90.51 in.: 64 in.:: DA : AC; 

and DA being yet unknown, we ascertain it by composition, thus: 

90.51 + 64 : 64 :: DA+AC, or DC : AC; 
or, 154.51 : 64 :: 64 : AC. 

Multiplying the middle terms of this proportion together, and divid¬ 
ing the product by the first term, we have the value of the last term, 
or AC, equal to 26.51 inches. Double this, and we have AB, the re¬ 
quired side , equal to 53.02 inches, or 4 ft. 5.02 in., and OC 2 +AC 2 = 
AO 2 , or 702.78+4096=4798.78 inches, the square root of which is 
69.27 inches, or 5 ft. 9.27 in.; and the whole of the required diagonal 
equals twice this number, or 11 ft. 6.54 in. 


Note .—Since all regular octagons are similar figures, any two regular octagons of dif¬ 
ferent dimensions will not only have their sides proportional, but their widths and their 
diagonal widths proportional also; and if we have the exact dimensions of all the parts 
of one octagon given, and any one part of the other octagon also given, then all its re¬ 
maining parts can be found by proportion. 

Example 1. Required the diagonal width of a regular octagon, the 
side of which is 12 feet. 

Let us compare this with another octagon, all the dimensions of 
which we know, or which we can find from the Table; say, an octagon 
of 16 feet side, the diagonal width of which, as given in the Table, is 


EXPLANATION OF TABLES. 137 

41 ft. 9.7 in.; then, since all the parts of the one figure are propoi- 
tional to the corresponding parts of the other, we shall have 
side to side, as diagonal width to diagonal width; 
or, 16 :12 :: 41 9.7 to the answer =31 ft. 4.27 in., 

Thus: 

AVe multiply the second and third terms together, and divide by the 
first: 

41 ft. 9.7 in. 

12 


492 

9.7 

16)501.7(31.856 feet ana decimals of a foot, which we reduce to 
48 feet and inches, thus: 


21 

31.356 

16 

12 

57 

sk27 

48 


90 


80 


100 



By multiplying the tenths of a foot by 12 to bring them to inches, 
and disregarding the third decimal figure, we have for a final answer 
31 ft. 4 t 2 o 7 u in. as the required answer; which is verified by the num¬ 
ber as given in the Table. 

Example 2. Required the side of a regular octagon, the square width 
of which is 30 feet. AVe compare this with the same octagon as be¬ 
fore, and have 

width to width as side to side; 

or, 88 7.52 : 30 :: 16 : to the answer =12 feet 5.12 inches, 

Thus: 

as in this case the first term or divisor is a compound number, we 
reduce all the three given terms to inches, and have 
* 88 ft. 7.52 in. =463.52 inches, 

30 feet =360 inches, 

16 feet =192 inches. 








138 CARPENTRY MADE EASY. 

So that the proportion, in inches, is 

463.52 : 360 :: 192 : answer =12 ft. 5.12 in. 

360 


11520 

576 


463.52)69120(149.119 in., or 12 ft. 5.12 in., Ana 
46352 


227680 

185408 


422720 

417168 


55520 

46352 


91680 

46352 


€53280 









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EXPLANATION OF TABLES, 


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TABLE IT 


The length of braces , like the length of rafters, must be determined 
by extracting the square root of the sum of the squares of the perpen¬ 
dicular and the horizontal runs. 

This table embraces almost every length that can be required in 
framing buildings, and comprises those of both regular and irregular 
runs. 

The exact length is here given to the hundredth part of an inch; 
but practically it will be found best to cut each brace from a sixteenth 
to an eighth of an inch longer than the exact rule requires, in order 
to compensate for compression and the shrinkage of the timber. 

( 140 ) 


EXPLANATION OF TABLES. 141 

TABLE IV. 

■Length of Braces given in Feet, Inches, and Hundredths of an Inch. 


f Length of Run. 

Length of Brace. 

Length of Run. 

Length of Brace. 

Length of Run. 

Length of Brace.. 

Ft. In. Ft. In . 


Ft. In. Ft. In. 


Ft. In. Ft. In. 


0 : 6 by 0 : 6 

0: 8.48 

3 : 3 by 3 : 3 

4: 7.15 

5 : 6 bv 5:6 

7: 9.33 

0 : 6 by 0 : 9 

0 : 10.81 

3 : 3 by 3 : 6 

4: 9.31 

5 : 9 by 5:9 

8: 1.58 

0 : 9 by 0 : 9 

1 : 0.72 

3 : 3 by 3 : 9 

4 : 11.54 

6 : 0 by 6:0 

*8: 5.82 

0 : 9 by 1 : 0 

1 : 3 

3 : 3 by 4 : 0 

5: 1.84 

6 : 3 by 6:3 

8 : 10.06 

1:0 by 1 : 0 

1 : 4.97 

3 : 6 by 3 : 6 

4 : 11.39 

6 : 6 by 6:6 

9 : 2.30 

1 : 0 by 1 : 3 

1: 7.20 

3 : 6 by 3 : 9 

5: 1.55 

6:9 by 6:9 

9 : 6.55 

i 1 : 3 by 1 : 3 

1 : 9.23 

3 : 6 by 4 : 0 

5: 3.78 

7 : 0 by 7:0 

9 : 10.79 

1 : 3 by # 1 : 6 

1 : 11.43 

3 : 9 by 3 : 9 

5: 3.63 

7 : 3 by 7:3 

10: 3.03 

1 : 6 by 1 : 6 

<2: 1.45 

3 : 9 by 4 : 0 

5 : 5.79 

7 : 6 by 7:6 

10: 7 28 

1 : 6 by 1 : 9 

2 : 3 6.1 

4 : 0 by 4 : 0 

5: 7.88 

7 : 9 by 7:9 

10 : 11.52 

1 : 9 by 1 : 9 

2: 5.69 

4 : 0 by 4 : 3 

5 : 10.03 

8 : 0 by 8:0 

11: 3.713 

1 : 9 by 2 : 0 

2 : 7.89 

4 : 0 by 4 : 6 

6 : 0.25 

8 : 3 by 8:3 

11: 8 

2 : 0 by 2 : 0 

2 : 9.94 

4 : 0 by 4 : 9 

6 : 2.51 

8 : 6 by 8:6 

12: 0.24 

2 : 0 by 2 : 3 

3: 0.12 

4 : 0 by 5 : 0 

6: 4.83 

8 : 9 by 8:9 

12: 4.49 

2 : 0 by 2 : 6 

3: 2.41 

4 : 3 by 4 : 3 

6: 0.12 

9 : 0 by 9:0 

12: 8.73 

2 : 3 by 2 : 6 

3: 4.36 

4 : 3 by 4 : 6 

6 : 2.27 

9:6 by 9:6 

13: 5.22 

2 : 6 by 2 : 6 

3: 6.42 

4 : 3 by 4 : 9 

6: 4.49 

10 : 0 by 10 : 0 

14: 1.70 

2 : 6 by 2 : 9 

3 : 8.59 

4 : 3 by 5 : 0 

6: 6 74 

10 : 6 by 10 : 6 

14 : 10.19 

2 : 9 by 2 : 9 

8 : 10.66 

4 : 6 by 4 : 6 

6 : 4.36 

11 : 0 by 11 : 0 

15: 6.67 

2 : 9 by 3 : 0 

4 : 0.83 | 

I 

4 : 6 by 4 : 9 

6: 6.51 

11 : 6 by 11 : 6 

1\1: 3.16 

3 : 0 by 3 : 0 

4: 2.91 

4 : 6 by 5 : 0 

6: 8.72 

12 : 0 by 12 : 0 

16: 11.64 

3 : 0 by 3 : 3 

4 : 5.02 

4 : 9 by 4 : 9 

6: 8.61 

12 : 6 by 12 : 6 

17 : 8.13 

3 : 0 by 3 : 6 

4: 7.31 

4 : 9 by 5 : 0 

6 : 10.75 

13 : 0 by 13 : 0 

18: 4.61 

3 : 0 by 3 : 9 

4: 9.62 

5 : 0 by 5 : 0 

7 : 0.85 

13 : 6 by 13 : 6 

19: 1.10 

3 : 0 by 4 : 0 j 

5 : 

5 : 3 by 5 : 3 

7: 5.09 

14 : 0 by 14 : 0 

19: 9.58 























































































































































































































142 


CARPENTRY MADE EASY, 


TABLE Y. 

Weight of Square Iron in Pounds and Ounces. 


Size. 

1 foot. | 

2 feet 


S feet. 

4 feet. 

6 feet 

• 

6 feet 


7 feet 


8 feet 


9 feet. 

Inches. 

lbs. 

oz. 

lbs. oz. 

lbs. oz. 

lbs. oz. 

lbs. oz. 

lbs. oz. 

lbs. oz. 

lbs. oz. 

lbs. oz. 

X 

Q: 

13 

1: 

11 

2: 

8 

2: 

6 

4: 

3 

5 : 

2 

5 : 

15 

6: 

13 

7: 

10 

% 

1: 

5 

2: 

10 

4: 

0 

5: 

5 

6: 

10 

7: 

9 

9 : 

5 

10 : 

10 

12: 

0 

% 

1 : 

15 

3: 

14 

5: 

13 

7 : 12 

9 : 

10 

11: 

9 

13 : 

7 

15 : 

6 

17: 

4 

X 

2: 

10 

5: 

4 

7: 

14 

10: 

8 

13: 

0 

15: 

8 

18: 

2 

20 : 

12 

23: 

6 

1 

3: 

6 

6: 

12 

10: 

2 

13: 

5 

16: 

12 

20 : 

4 

23 : 

10 

27: 

0 

30: 

6 

1 A 

4: 

5 

8: 

10 

12: 

15 

17: 

3 

21: 

8 

25 : 

12 

30 : 

0 

34: 

5 

38 : 

9 

IX 

5 : 

5 

10: 

10 

15: 

15 

21: 

2 

26: 

7 

31: 

12 

37: 

0 

42: 

4 

47: 

8 

1 X 

6 : 

6 

12: 

12 

19 : 

3 

25: 

9 

32: 

0 

38 : 

6 

44 : 

12 

51: 

2 

57: 

8 

IX 

7 : 

10 

15: 

4 

22: 

12 

30: 

6 

38 : 

0 

45: 

10 

53 : 

4 

60: 

14 

68 : 

7 

ix 

8 : 

14 

17 : 

12 

26: 

11 

35 : 

9 

44: 

8 

53: 

7 

62: 

6 

71 : 

5 

80 : 

4 

1% 

10: 

6 

20: 

12 

31: 

2 

41 : 

7 

51: 

13 

62: 

2 

72: 

8 

82: 

14 

93: 

4 

ix 

11 : 

14 

23 : 

12 

35 : 

10 

47 : 

8 

59 : 

6 

71: 

5 

83 : 

3 

95: 

2 

107 : 

0 

2 

13 : 

8 

27 : 

0 

40 : 

10 

54 : 

2 

67 : 

10 

81 : 

2 

94: 

10 

102: 

03 

121 : 

11 

2X 

15 : 

5 

30 : 

8 

45 : 

13 

61: 

2 

76 : 

6 

91: 

10 

106 : 

13 

122: 

2 

137 : 

6 

2X 

17 : 

2 

34: 

3 

51 : 

5 

68 : 

6 

85: 

9 

102: 

11 

119 : 

13 

136 : 

15 

154: 

0 

2X | 

19 : 

1 

38 : 

2 

57: 

3 

76: 

4 

95: 

5 

114: 

6 

133 : 

7 

152: 

8 

171 : 

9 

2X 

21 : 

1 

42 : 

3 

63: 

6 

84: 

8 

105 : 

10 

126 : 

14 

147 : 

15 

169 : 

0 

190: 

1 

2X 

23 : 

5 

46 : 

10 

69 : 

15 

93 : 

3 

116 : 

8 

139 : 

13 

163 : 

i 

186 : 

5 

209 : 

10 

2% 

25 : 

10 

51 : 

02 

76: 

12 

102: 

3 

127 : 

13 

153 : 

6 

178 : 

15 

204 : 

8 

230: 

0 

2X 

27 : 

14 

65 : 

14 

83 : 

13 

111 : 

12 

139 : 

11 

167 : 

10 

195 : 

9 

223 : 

8 

251 : 

7 

3 

30 : 

6 

60 : 

12 

91 : 

3 

121: 

9 

152: 

2 

1S2: 

8 

212: 

14 

243: 

5 

273: 

11 

3X 

33 : 

0 

66 : 

0 

99 : 

0 

132: 

0 

165: 

1 

198 : 

1 

231 : 

1 

264 : 

2 

297 : 

2 

3X 

35 : 

11 

71 

6 

107 : 

2 

142: 

13 

178: 

8 

214: 

3 

249 : 

15 

2S5: 

10 

321 : 

5 

3X 

38 : 

8 

77 : 

0 

115 : 

8 

154: 

0 

192: 

8 

231: 

0 

269 : 

8 

308 : 

0 

346 : 

8 

8X 

41 : 

6 

S2 : 

12 

124: 

3 

167 : 

9 

207 : 

0 

248 : 

6 

289 : 

12 

331 : 

3 

372 : 

10 

3X 

44: 

6 

88 : 

12 

133 : 

4 

177 : 

11 

221: 

2 

266 : 

8 

310 : 

14 

355: 

5 

399 : 

13 

3X 

47 : 

8 

95 : 

1 

142 : 

9 

190: 

2 

237 : 

11 

285 : 

3 

332: 

11 

380 : 

5 

427 : 

13 

3X 

50 : 

13 

101: 

8 

152: 

5 

203: 

0 

253: 

13 

304: 

8 

355 : 

5 

406: 

0 

456 : 

13 

4 

54 : 

2 

108 : 

3 

162: 

4 

216: 

5 

270 : 

6 

324: 

8 

376 : 

10 

432 : 

11 

486 : 

13 

4 X 

57 : 

8 

115 

0 

172: 

10 

230: 

2 

287 : 

10 

345 : 

2 

402: 

10 

460 : 

2 

517 : 

11 

4 >4 

61 

1 

122: 

2 

183 : 

3 

244 : 

4 

305 : 

5 

366 : 

6 

427: 

7 

4SS : 

7 

549 : 

8 

4X 

64 

11 

129 

6 

194: 

2 

258 : 

13 

323 : 

8 

388: 

3 

452: 

14 

517 

10 

582: 

5 

4X 

68 

6 

136 

14 

205 

5 

273 : 

13 

342: 

3 

410: 

11 

479 

2 

647: 

10 

616 : 

0 

4X 

72: 

5 

144 

10 

216 

15 

2S9 : 

3 

361 : 

8 

433 

13 

506: 

2 

578: 

7 

450 : 

11 

4X 

76 

5 

152 

8 

228 

12 

305 : 

1 

381 

6 

457 

9 

533 

13 

610 

1 

686 : 

6 

4X 

80: 

5 

160 

11 

241 

0 

221 : 

5 

401 

11 

482 

0 

562 

5 

642 

11 

723 : 

0 

6 

84 

8 

169 

0 

253 

6 

337 : 

15' 

422: 

6 

606 

15 

591 

6 

675 

14 

760: 

5 






























































































































































































































































































































































































































































































EXPLANATION OF TABLES, 


143 


TABLE V.— Continued. 


Weight of Square Iron in Pounds and Ounces. 


Size. ! 

10 feet. 

11 feet. 

12 feet. 

13 feet. 

14 feet. 

16 feet. 

16 feet. 

17 feet. 

18 feet. 

Inches. 

lbs. 

oz. 

lbs. oz. 

lbs. 

OZ. 

lbs. oz. 

lbs. oz. 

lbs. oz. 

lbs. 

OZ. 

lbs. oz. 

lbs. oz. 

M 

8 : 

s ! 

9 : 5 

10 : 

2 

11 : 0 

11 : 13 

12: 9 

13: 

8 

14: 6 

15: 3 

X 

13: 

5 

14 : 10 

16 : 

0 

17: 5 

18 : 10 

19: 15 

21 : 

4 

22: 9 

23 :14 

X j 

19 : 

0 

20 : 15 

22: 

13 

24 : 12 

26 :10 

28: 8 

30: 

6 

32: 4 

34: 02 

% ! 

26 : 

0 

28: 8 

31 : 

2 

33 : 12 

36: 6 

38 : 14 

41 : 

8 

44: 2 

46 : 17 

i 

33 : 

12 

37 : 3 

40 : 

10 

44: 0 

47: 6 

50 : 12 

54: 

2 

57: 8 

60: 14 

ix 

42 : 

14 

47 : 2 

51 : 

6 

55 : 11 

59 : 15 

64: 4 

68: 

8 

72: 13 

77: 0 

lx 

52: 

12 

5S: 1 

63 : 

6 

68 : 11 

74: 0 

79: 4 

84: 

8 

89 : 13 

95: 0 

i x 

63 : 

14 

70: 4 

76: 

10 

83: 0 

89 : 7 

95 : 14 

102: 

2 

108: 9 

115: 0 

IX 

76: 

0 

83 : 10 

91 : 

4 

9S : 14 

106: 8 

114: 2 

121 : 

12 

129 : 6 

137; 0 

ix 

89 : 

4 

98: 3 

107 : 

2 

116: 1 

125: 0 

133 : 15 

142 : 

14 

151 : 13 

160 : 11 

ix 

103 : 

10 

114: 0 

124: 

5 

134 : 11 

145: 0 

155: 5 

165 : 

11 

176: 0 

186: 6 

IK 

118 : 

13 

130 : 11 

142: 

10 

154: 8 

166: 6 

178: 3 

IStO: 

2 

202: 0 

213 : 14 

2 

135 : 

3 

14S : 11 

162 : 

3 

175 : 12 

189: 4 

202: 13 

216: 

5 

229 : 13 

243: 6 

2K 

152 : 

10 

167 : 14 

183: 

3 

198: 7 

213 : 11 

229 : 0 

244: 

5 

259 : 9 

274: 14 

2K 

171 : 

2 

188: 3 

205 : 

5 

222: 8 

239 : 10 

256: 11 

273: 

13 

290 : 15 

308 : 0 

2K 

190 : 

10 

209 : 11 

228 : 

12 

247 : 14 

266 : 15 

*286 : 0 

305 : 

1 

324 : 2 

343: 3 

2K 

211 : 

3 

232: 5 

253: 

7 

274 : 9 

295 : 12 

316 : 14 

337 : 

15 

359 : 0 

380 : 3 

2K 

232 : 

15 

256 : 3 

279 : 

8 

302 : 13 

326: 2 

349 : 7 

372: 

12 

396 : 0 

419: 5 

2K 

255 : 

10 

281 : 3 

306 : 

13 

332: 6 

357 : 13 

383 : 6 

409: 

434: 8 

460: 1 

2K 1 

279 : 

6 

307 : 5 

335 : 

4 

363 : 3 

391 : 1 

419: 0 

447 : 

0 

475 : 0 

502: 14 

3 

304 : 

2 

334: 9 

365 : 

0 

395 : 6 

425 : 13 

456 : 3 

486: 

10 

517: 1 

547 : 8 

3X 

330 : 

2 

363: 2 

396 : 

2 

329 : 2 

462: 2 

495 : 3 

528: 

3 

561: 3 

594: 3 

3K 

357 : 

0 

392 : 11 

428: 

6 

464 : 3 

500 : 0 

535 : 11 

571: 

5 

607 : 0 

642: 11 

3K 

385 : 

0 

423 : 8 

462: 

0 

500 : 8 

539 : 0 

577: 8 

616 : 

0 

654 : 9 

693: 1 

3K 

414 : 

1 

455 : 8 

496 : 

15 

538 : 5 

579 : 11 

621 : 2 

662: 

8 

703 : 14 

745 : 5 

3K 

444 : 

3 

488: 9 

533 : 

0 

577: 6 

621 : 15 

666 : 5 

710: 

11 

755 : 2 

799 : 8 

3K 

475 : 

5 

522 : 14 

570 : 

6 

617 : 15 

665: 8 

713: 0 

760 : 

8 

808: 1 

855: 10 

3K 

507 : 

10 

558: 5 

609 ; 

1 

659 : 13 

710 : 10 

761: 5 

812: 

2 

862 : 15 

913: 10 

4 

540: 

14 

594: 15 

649 : 

0 

703 : 2 

757 : 3 

811: 4 

865: 

5 

9] 9 ; 6 

973 : 8 

4K 

575 : 

3 

632 : 11 

690 : 

3 

747 : 11 

805 : 3 

862 : 12 

920 : 

5 

977 : 13 

1035: 5 

4K 

610 : 

9 

671 : 10 

732 : 

11 

793 : 12 

854 : 13 

915 : 14 

976 : 

15 

1037 : 15 

1099 : 0 

4K 

646 : 

0 

711 : 11 

776 : 

6 

841 : 2 

905 : 13 

970 : 8 

1035: 

3 

1099 : 14 

1164: 10 


684 : 

8 

752 : 15 

821 : 

6 

8S9 : 13 

958 : 5 

1026: 11 

1095 : 

3 

1163 : 10 

1232 : 2 

4K 

721: 

1 

795 : 6 

867 : 

11 

940 : 0 

1012: 5 

1084 : 10 

1156: 

15 

1229 : 3 

1301 : 8 

4K 

762 : 

10 

838 : 15 

915 : 

3 

991 : 7 

1067 : 11 

1144: 0 

1220: 

3 

1296: 8 

1372: 13 

4K 

803 : 

5 

8S3 : 11 

964: 

0 

1044 : 5 

1124 : 11 

1205: 0 

1285 : 

5 

1365 : 11 

1446: 0 

5 

844: 

13 

929 : 5 

1013: 

14 

1098: 3 

1J82: 11 

12i»7: 3 

1351 : 

11 

1436: 3 

1520 : 10 





























































































































































































































































































































































































































































































144 


CARPENTRY MADE EAS Y 


TABLE YI. 

Weight of Flat Iron in Pounds and Ounces. 


Thick. 

Wide, j 

X foot. | 

2 feet. 

8 feet. 

4 feet. 

5 feet. 

6 feet. 

7 feet. 

8 feet. 

9 feet. 

11 feet. 

Inches. 

Inches. 

lbs. 

oz. 

lbs. 

OZ. 

lbs. 

OZ. | 

lbs. 

OZ. 

lbs. 

OZ. 

lbs. 

OZ* 

lbs. 

OZ. 

lbs. 

OZ. 

lbs. 

OZ. 

lbs. 

01 . 


1 | 

0: 

13 

1 : 

1 

11 

2: 

8 

3: 

6 | 

4 : 

2 

5 : 

0 

5 : 

14 

6 : 

13 

7 : 

10 

8 : 

8 


ix | 

1: 

1 

2 : 

2 

. 3: 

-- 

3 

4 : 

3 

5 : 

4 

6 : 

5 

7 : 

6 

8 : 

7 

9 : 

8 

10: 

9 


IK 

1: 

4 

2: 

8 

3 : 

12 

5: 

1 

6: 

5 

7 : 

9 

8 : 

13 

10 : 

i ! 

i 

11 : 

5 

12: 

10 ( 


IK 

1 : 

8 

3 : 

0 

4: 

7 

5: 

14 

7 : 

6 

8 : 

14 

10 : 

6 

11 : 

13 

13: 

5 

14: 

13 


2 


11 

3 : 

6 

5 : 

1 

6 : 

13 

S: 

8 

10: 

1 

11 : 

13 

13: 

8 

15 : 

3 1 

16 : 

14 


2 X 

1 : 

14 

3 : 

12 

5 : 

11 

7 : 

10 

9: 

8 

11 : 

6 

13: 

4 

15 : 

3 

17 : 

1 

19: 

0 

k- 

2 X 

2: 

1 

4 : 

3 

6 : 5 

8: 

7 

10 : 

9 

12 : 

11 

14: 

13 

16 : 

14 

19 : 

0 

21: 

1 


2 K 

2: 

5 

4 : 

10 

7 : 

0 

9 : 

5 

11 : 

10 

13 : 

14 

16 : 

5 

IS : 

10 

20 : 

15 

22: 

3 


3 

2 : 

8 

5 : 

1 

7 : 

10 

10 : 

1 

12 : 

11 

15: 3 

17 : 

11 

20: 

5 

22 : 

13 

25 : 

5 


3K 

2: 

12 

5 : 

8 

8: 

3 

11 : 

0 

13: 

12 

16: 

8 

19 : 

3 

22: 

0 

24 : 

12 

27 : 

8 


3K 1 

3 : 

0 

5 : 

14 

8 : 

13 

11 : 

12 

14: 

11 

17 : 

10 

20 : 

10 

23 : 

9 

26 : 

9 

29: 

8 


3M j 

3 : 

3 

6 : 

5 

9 : 

8 

12 : 

11 

15 : 

13 

19 : 

0 

22: 

3 

25: 

6 

28: 

8 

31 : 

11 


4 

3: 

6 

6 : 

13 

10 : 

2 

13: 

8 

16 : 

14 

20: 

5 

23 : 

11 

27 : 

0 

30 : 

6 

33 : 

13 


1 

1 : 

4 

2 : 

8 

3 : 

12 

5 : 

1 

6 : 

5 

7 : 

9 

8 : 

14 

10 : 

2 

11 : 

6 

12: 

11 


IK 

1 : 

10 

3 : 

4 

4 : 

13 

6 : 

5 

7 : 

14 

9 : 

8 

11 : 

1 

12 : 

11 

14: 

5 

15: 

13 


IK | 

1 : 

14 

3 : 

12 

5 : 

11 

• 7 : 

10 

9 : 

8 

11 : 

6 

13 : 

5 

15 : 

2 

17 : 

1 

19: 

0 


IK 

2: 

3 

4 : 

6 

6: 

10 

8 : 

14 

11 : 

1 

13 : 

5 

15 : 

8 

17 : 

11 

20 : 

1 

22: 

3 


2 

2: 

8 

5 : 

1 

7 : 

9 

10 : 

1 

12: 

9 

15 : 

2 

17 : 

11 

20 : 

5 

22: 

11 

25 : 

6 


2K 

2 : 

14 

5 : 

11 

8 : 

8 

11: 

6 

14 : 

4 

17 : 

2 

20 : 

0 

22 : 

12 

25 : 

10 

28: 

8 


2K 

3 : 

3 

6 : 

0 

9 : 

8 

12: 

11 

15: 

13 

19 : 

0 

22: 

3 

25 : 

6 

28 : 

8 

31 : 

11 


2K 

3 : 

8 

7 : 

0 

10 : 

8 

13 : 

15 

17: 

7 

20 : 

14 

24: 

7 

27 : 

15 

31 : 

7 

34: 

15 


3 

3: 

13 

7 : 

10 

11 : 

6 

15: 

3 

19 : 

0 

22 : 

12 

26 : 

9 

30: 

6 

34 : 

3 

38 : 

0 


3K 

4: 

2 

8 : 

4 

12: 

6 

16: 

8 

20: 

10 

24: 

12 

28 : 

14 

33 : 

0 

37 : 

2 

41 : 

4 


3K 

4: 

7 

8 : 

14 

13: 

5 

17 : 

12 

22 : 

3 

26: 

10 

31 : 

0 

35 : 

7 

39 : 

15 

44 : 

6 


3K 

4 : 

12 

9 : 

8 

14 : 

4 

19 : 

0 

23 : 

12 

28 : 

8 

33 : 

4 

38 : 

0 

42 : 

12 

47 : 

8 


4 

5 : 

1 

10: 

2 

15 : 

3 

20: 

4 

25 : 

5 

30: 

6 

35 : 

8 

40: 

9 

45 : 

10 

50 : 

11 


1 

1: 

11 

3 : 

6 

5 : 

1 

6 : 

13 

8: 

8 

10: 

2 

11 : 

13 

13: 

8 

15 : 

3 

16 : 

14 


IK 

2: 

1 

4 : 

3 

6 : 

5 

8: 

7 

10 : 

9 

12: 

11 

14: 

13 

16: 

14 

19 : 

0 

21 : 

2 


IK 

2: 

8 

5 : 

1 

7 : 

9 

1 10: 
l 

1 

12: 

10 

15 : 

3 

17 : 

11 

20: 

5 

22 : 

13 

25 : 

5 


IK 

3 : 

0 

5 : 

15 

8: 

14 

11 : 

13 

14: 

12 

17: 

11 

I 20: 

10 

23: 

1C 

26: 

9 

1 29 : 

9 


2 

3 : 

6 

6: 

12 

10: 

2 

13 : 

8 

16: 

14 

1 20: 

5 

23 : 

10 

27 : 

0 

30 : 

6 

33: 

12 


2K 

3 : 

13 

7: 

10 

11 : 

7 

15: 

3 

19 : 

0 

22: 

13 

26 : 

10 

30 : 

7 

34: 

4 

38 : 

0 


2K 

4 : 

3 

8: 

6 

12: 

10 

16 : 

14 

21 : 

2 

25 : 

5 

26 : 

9 

33 : 

14 

38 : 

2 

42: 

4 


2K 

1 4: 

10 

9 : 

5 

13 : 

15 

| 18 : 

9 

23: 

3 

27 : 

14 

32 : 

8 

37 : 

3 

41 : 

13 

; 46: 

8 


3 

5 : 

1 

10 : 

2 

15 : 

3 

20 : 

4 

25: 

5 

30: 

6 

35: 

8 

40 : 

9 

45 : 

10 

50 • 

11 


3K 

1 5 : 

8 

11 : 

0 

16 : 

8 

22 : 

0 

27 : 

8 

33 : 

0 

38 : 

8 

44 : 

0 

49 : 

8 

55 : 

0 

i 

3K 

! 5 : 

14 

11 : 

12 

1 17: 

11 

23 

10 

29 : 

9 

1 35: 

8 

41 : 

7 

47 : 

5 

53 : 

4 

59 : 8 





























































































































































































































































































































































































































































































































































































































EXPLANATION OF TABLES. 145 

TABLE YI.— Continued . 


Weight of Flat Iron in Pounds and Ounces. 


Thick. 

Wide. 

1 foot. 

2 feet. 

3 feet. | 

4 feet. 

5 feet. 

6 feet. 

7 feet. 

8 feet. 

9 feet. 

10 feet. 

laches. 

Inches. 

lbs. oz. 

lbs. oz. 

lbs. oz. 

lbs. oz. 

lbs. oz. 

lbs. oz. 

lbs. oz. 

lbs. oz. 

lbs. oz. 

lbs. o*. 


<- 

3X 

6: 5 

12: 11 

19: 0 | 

25 : 5 

31 : 11 

38: 0 

44: 6 

50 : 11 

57: 0 

63: 5 

w 

r 

4 

6 : 12 

13: 8 

20: 5 

27: 0 

33 : 12 

40: 9 

47: 4 

54 : 0 

60: 12 

67: 9 



1 

2: 1 

4: 3 

6: 5 

8: 7 

10: 9 

12 :11 

14 : 13 

16 : 15 

19: 0 

21: 2 



IX 

2 : 10 

5: 5 

7 : 15 

10: 9 

13; 3 

15 : 13 

18: 8 

21 : 2 

23: 12 

26 : 6 



ix 

3: 3 

6: 6 

9: 8 

12: 11 

15 : 13 

19: 0 

22: 3 

25 : 6 

28: 8 

31 : 11 



IX 

3: 11 

7: 6 

11 : 1 

14 : 13 

18: 8 

22: 3 

25: 14 

29: 9 

33: 4 

37: 0 



2 

4: 3 

8: 6 

12 : 10 

16 : 14 

21 : 1 

25: 5 

29: 9 

33 : 12 

38: 0 

42: 3 



2X 

4 : 12 

9: 8 

14: 5 

19: 0 

23 : 12 

28: 8 

33: 5 

38: 0 

42 : 12 

47 : 8 



2X 

5 : 5 

10: 9 

15 : 13 

21 : 1 

26: 6 

31 : 10 

37: 0 

42: 3 

47 : 8 

52 : 12 



2X 

5 : 13 

11:10 

17: 6 

23 : 3 

29: 0 

34 : 13 

40 : 11 

46: 8 

52: 5 

58: 2 



3 

6 : 5 

12 : 11 

19: 0 

25 : 5 

31 : 11 

38: 0 

44: 6 

50 : 11 

57 : 10 

63: 0 



3X 

6 : 14 

13 : 12 

20: 9 

27 : 8 

34: 5 

41 : 3 

48: 1 

55: 0 

61 : 12 

68 : 10 



3X 

7: 6 

14 : 12 

22: 3 

29: 9 

37: 0 

44: 6 

51 : 13 

59: 3 

66: 8 

73: 14 



3X 

7 : 14 

15 : 12 

23 : 11 

31 : 10 

39: 9 

47: 8 

55 : 7 

63: 6 

71: 5 

79: 4 



4 

8: 7 

16: 14 

25: 5 

33 : 12 

42: 3 

50 : 10 

59: 1 

67: 9 

76: 0 

84: 8 



1 

2: 8 

5: 1 

7 : 9 

10: 1 

12 : 10 

15: 3 

17 : 11 

20: 5 

22 : 13 

25: 5 



lx 

3: 3 

6 : 6 

9 : 8 

12 : 10 

15: 13 

19: 0 

22: 3 

25: 6 

28: 8 

31 : 11 



ix 

3 : 12 

7: 9 

11: 6 

15: 3 

19: 0 

22: 12 

26: 9 

30: 6 

34: 3 

38: 0 



ix 

4: 7 

8 : 14 

13: 5 

17 : 12 

22: 3 

26 : 10 

31 : 1 

35 : 8 

39 : 15 

44: 6 



2 

5: 1 

10: 2 

15: 3 

20- 4 

25: 5 

30: 6 

35: 8 

40: 9 

45: 10 

50 : 11 



2X 

5 : 11 

11: 6 

17: 1 

22 : 12 

28: 8 

34: 3 

39 : 15 

45 : 10 

51: 5 

■ 57: 0 



2X 

6: 5 

12: 10 

19: 0 

25 : 5 

31 : 10 

38: 0 

44: 6 

50 : 11 

57: 0 

63: 5 



2X 

7: 0 

14: 0 

21 : 0 

28: 0 

35 : 0 

42: 0 

49 : 0 

56: 0 

63: 0 

69: 0 



3 

7: 10 

15: 4 

22 : 13 

30 : 6 

38: 0 

45 : 10 

53: 4 

60 : 14 

68: 8 

76: 0 



3X 

8: 4 

16: 8 

24: 12 

33: 0 

41 : 4 

49 : 7 

57 : 12 

66: 0 

74: 3 

82: 7 



3X 

8: 14 

17 : 12 

26 : 10 

35 : 8 

44: 6 

53: 4 

62: 6 

71: 0 

79 : 14 

88 : 12 



3X 

9 : 8 

19: 0 

28 : 8 

38: 0 

47: 8 

57: 0 

66: 8 

76: 0 

85: 8 

95 : 0 



4 

10: 2 

20: 4 

30: 6 

40: 9 

50 : 11 

60 : 14 

70 : 15 

81 : 1 

91: 3 

101: 6 


** 

IX 

5: 1 

10: 2 

15: 3 

20: 4 

25 : 5 

30: 6 

35: 8 

40: 9 

45 : 10 

50 : 11 



2 

6 : 12 

13: 8 

20: 4 

27: 0 

33: 12 

40: 9 

47: 5 

54: 0 

60 : 12 

67: 9 

i- 


3 

10: 2 

20: 4 

30: 6 

40: 9 

50 : 11 

60: 13 

70 : 15 

81: 1 

91 : 3 

101 : 5 


4 

13: 8 

27: 0 

40: 9 

54: 1 

67: 9 

81 : 1 

94 ^ 9 

108: 1 

121 : 10 

135: 2 



6 

16: 14 

33: 12 

50: 11 

67 : 9 

84: 8 

101 : 6 

118: 5 

135: 4 

152: 2 

169 : O 



6 

20: 5 

40: 10 

60: 15 

81: 2 

101: 6 

121 : 11 

141 : 14 

162: 3 

182 : 8 

202 : 12 


10 


















































































































































































































































































































































































































































































146 CARPENTRY MADE EASY. 

TABLE VII. 

Weight of Round Iron in Pounds and Ounces. 


Size. 

1 foot. 

2 feet. 

3 feet. 

4 feet, j 

6 feet. 

6 feet. 

7 feet. 

8 feet. 

9 feet. 

Diameter 
in inches. 

Ibs. oz. 

lbs. oz. 

lbs. oz. 

lbs. oz. 

lbs. oz. 

lbs. oz. 

lbs. oz. 

lbs. oz. 

lbs. oz. 

y* 

0 : 11 

1: 5 

2: 0 

2 :11 

3: 5 

4: 0 

4 : 11 

5: 5 

6: 0 

% \ 

1: 0 

2: 1 

3: 2 

4: 3 

5: 4 

6: 5 

7: 5 

8: 6 

9: 7 

X 

1 : 8 

3: 0 

4: 8 

6: 0 

7 : 8 

9 : 0 

10: 8 

12: 0 

13: 8 
- 1 

% 

2: 0 

4: 1 

6: 1 

8: 2 

10: 3 

12: 3 

14: 4 

16 : 5 

18: 5 

i 

2: 11 

5: 5 

8: 0 

10 : 10 

13: 5 

15 : 15 

18 : 10 

21 : 3 

23: 14 

ix 

3 : 6 

6 : 12 

10: 2 

13: 7 

16 : 13 

20: 3 

23: 8 

26 : 14 

30: 4 

ix 

4: 3 

8: 6 

12: 8 

16 : 11 

20 : 14 

25 : 0 

29: 3 

33: 6 

37: S 


5 : 0 

10 : 0 

15: 1 

20: 1 

25: 2 

30 : 2 

35: 2 

40: 3 

45: 3 

1X 

6: 0 

11 : 15 

17 : 15 

23 : 14 

29 : 14 

35 : 13 

41 : 12 

47: 12 

53 : 11 


7: 0 

14: 0 

21 : 0 

28: 0 

35: 1 

42: 1 

49 : 1 

56: 1 

63: 1 

ix 

8: 2 

16: 4 

24: 6 

32: 8 

40 : 10 

48 : 12 

56 : 14 

65 : 0 

72: 3 

ix 

9: 5 

18 : 11 

28 : 

37 : 5 

46 : 11 

56 t 0 

65 : 5 

74 : 11 

84: 0 

2 

10 : 10 

21 : 4 

31 : 14 

42: 8 

53: 2 

63 : 12 

74: 5 

84: 14 

95: 8 

1 2X 

12: 

24: 

36 : 

48 : 

60: 

72: 

84 : 

96 : 

108: 

) 2)£ 

13: 8 

26 : 14 

40 : 6 

53 : 13 

67: 4 

80: 9 

94 : 2 

107 : 8 

121 : 

( 2X 

15 : 

30 : 

45 : 

60 : 

75: 

89 : 15 

104 : 13 

119 : 13 

134 : 13 

2>^ 

16 : 11 

33: 4 

50: 1 

66 : 12 

83: 7 

100 : 1 

116 : 12 

133 : 8 

150: 3 

2 X 

18 : 13 

37 : 10 

54: 6 

73 : 3 

92 : 10 

112: 8 

131 : 4 

150 : 0 

168 : 12 

2X 

20: 1 

40 : 3 

60: 5 

80: 6 

100 : 7 

120: 8 

140: 9 

160 : 10 

180: 11 

2X 

21 : 15 

43: 14 

65 : 13 

87 : 12 

109 : 11 

131 : 10 

153: 9 

175: 8 

197 : 7 

3 

23 : 14 

47 : 12 

71 : 11 

95 : 9 

119 : 7 

143: 5 

167: 3 

191 : 1 

215: 0 

3X 

25 : 14 

51 : 13 

77 : 12 

103 : 11 

129 : 10 

155 : 9 

181 : 8 

207 : 7 

233 : 6 

3X 

28: 0 

56: 1 

84: 2 

112: 3 

140: 3 

168: 4 

196: 5 

224: 5 

253 : 6 

3X 

30: 4 

60: 5 

90 : 12 

121 : 0 

151 : 4 

1S1 : 7 

211 : 11 

241 : 15 

272 : 3 

* 

CO 

32: S 

65: 0 

97: 8 

130: 0 

162 : 9 

195: 1 

127 : 9 

260 : 1 

292: 9 

3X 

34 : 14 

69 : 12 

104 : 10 

139: 8 

174: 6 

209 : 5 

244: 3 

279 : 1 

314: 0 

3X 

37 : 5 

74 : 11 

112 : 0 

149 : 5 

1S6 : 11 

224: 0 

261 : 5 

298 : 11 

336: 0 

3X 

39 : 14 

79 : 12 

119 : 10 

159: 8 

199 : 5 

239 : 3 

279: 0 

318 : 14 

358 : 12 ' 

4 

42: 8 

84 : 15 

127: 6 

169 : 14 

213 : 5 

254 : 13 

297 : 4 

339 : 12 

382: 4 ( 

4X 

45: 3 

90 : 5 

135 : 8 

ISO : 11 

22 ") • 14 

271 : 0 

316 : 4 

361 : 7 

406 : 10 

*X 

48: 0 

95 : 15 

143 : 14 

191 : 13 

239 : 12 

2S7 : 11 

335 : 10 

383: 9 

431 : 9 

4X 

50: 12 

101 : 11 

152: 7 

| 203 : 5 

254: 1 

304 : 15 

355 : 11 

406 : 8 

457 : 5 

*X 

53 : 12 

107 : S 

161 : 5 

215: 0 

268 : 12 

322 : 11 

376 : 5 

430: 2 

483 : 13 


56 : 12 

113 : 10 

170 : 6 

227 : 3 

283 : 15 

310 : 11 

397: 8 

451: 5 

511: 2 

4X 

60 : 0 

119 : 14 

179 : 12 

239 : 10 

299: 8 

359 : 6 

419 : 5 

479 : 2 

539: 1 

4X 

63: 2 

126: 3 

189 : 5 

252 : 6 

315: 8 # 

378 : 10 

441 : 11 

504 : 12 

567 : 13 

5 

- 

66 : 12 

133: 8 

200: 5 

267 : 0 

333 : 12 

400 : 8 

467 : 5 

534 : 0 

600 : 12 \ 






















































































































































































































































































































































































































































































































EXPLANATION OF TABLES, 


147 


TABLE YI1.— Continued. 


"Weight of Round Iron in Pounds and Ounces. 


Mize. 

19 feet. | 

11 feet. 

12 feet. 

13 feet. 

14 feet. 

15 feet 


16 feet 


17 feet. 

18 feet 

• 

Diameter 
j*x inches. 

lbs. 

oz. 

lbs. oz. 

lbs. 

3Z. 

lbs. 

OZ. 

lbs. 

OZ. 

lbs. oz. 

lbs. oz. 

lbs. oz. 

lbs. ox. 

y 

6: 

11 i 

7: 

5 

8 : 

0 

8 

11 

9 : 

5 

10 : 

0 

10: 

11 

11: 

5 

11 : 

14 

% 

10 : 

7 1 

11 : 

8 

12 : 

8 

13 

9 

14 : 

10 

15 : 

10 

16 : 

11 

17 : 

12 

IS: 

13 

X 

15 : 

0 

16 : 

8 

IS: 

0 

19 

8 

21 : 

0 

22: 

8 

24 : 

0 

25 : 

8 

27: 

0 

% 

20: 

6 

22 : 

7 

24: 

7 

26 

7 

28: 

8 

30 : 

8 

32 : 

8 

34: 

9 

36 : 

10 

i 

25 : 

8 

29 : 

3 

31 : 

13 

34 

8 

37 : 

3 

39 : 

13 

42: 

8 

45 : 

2 

47 : 

12 


33 : 

9 

37 : 

0 

40 : 

5 

43 

11 

47 : 

0 

50 : 

6 

53 : 

12 

57 : 

2 

60 : 

8 

1 

lx 

41 : 

11 

45 : 

14 

50 : 

1 

54 

4 

58: 

6 

62 : 

9 

66 : 

12 

70 : 

14 

75 : 

1 

1% 

50 : 

3 

55 : 

4 

60 : 

4 

65 

4 

70 : 

6 

75 : 

6 

80 : 

5 

85 : 

5 

90 : 

4 

iy 

59 : 

11 

65 : 

11 

71 : 

10 

77 

10 

83 : 

9 

89 : 

9 

95 : 

10 

101 : 

8 

107 : 

8 


70 : 

2 

77 : 

2 

84: 

2 

91 

: 2 

98: 

3 

105 : 

3 

112 : 

3 

119 : 

3 

126 : 

3 

ix 

81 : 

5 

89 : 

7 

97 : 

9 

105 

: 11 

113 : 

13 

121 : 

15 

130 : 

0 

138 : 

2 

146 : 

4 

IVs 

93: 

5 

102: 

11 

112: 

0 

121 

: 5 

130: 

11 

140 : 

0 

149 : 

5 

158 : 

11 

168 : 

0 

2 

106 : 

2 

116 : 

12 

127 : 

6 

138 

: 0 

148 : 

10 

159 : 

4 

169 : 

14 

180 : 

8 

192 : 

2 

2X 

120: 

132: 

144: 

156 : 

168 : 

180 : 

192: 

0 

204 : 

0 

216 : 

0 

%X 

134 : 

7 

147 : 

13 

161 : 

5 

174 

: 11 

188 : 

2 

211 : 

10 

215 : 

0 

228 : 

8 

242: 

0 

2% 

149 : 

12 

164 : 

12 

179 : 

12 

194 

: 11 

209 : 

11 

224 : 

11 

239 : 

10 

254 : 

10 

269 : 

10 

2 y 2 

166 : 

14 

183 : 

9 

200 : 

4 

216 

: 15 

233 : 

10 

250 : 

5 

267 : 

0 

283 : 

10 

300 : 

5 

2% 

187 : 

8 

206 : 

4 

225 : 

0 

243 

: 12 

262 : 

8 

281 : 

4 

299 : 

11 

318 : 

12 

337 : 

8 

2^ 

200 : 

12 

220 : 

13 

240 : 

15 

261 

: 1 

281 : 

1 

301 : 

2 

321 : 

3 

341 : 

5 

361: 

6 

2^ 

219 : 

6 

241 : 

5 

263 : 

5 

285 

: 4 

307 : 

2 

329 : 

2 

351 : 

2 

373: 

0 

395 : 

0 

3 

238 : 

14 

262 : 

12 

286 : 

10 

310 

: 8 

334 : 

6 

35 S : 

4 

383: 

3 

406 : 

1 

430 : 

0 

3^ 

259 : 

5 

2S5 : 

4 

311 : 

3 

337 

: 1 

363 : 

0 

388 : 

14 

414 : 

12 

440 : 

11 

466 : 

10 

3^ 

280 : 

7 

308 : 

7 

336 : 

8 

364 

: 8 

392 : 

9 

320 : 

10 

448 : 

10 

476 : 

11 

504 : 

11 

3% 

302 : 

7 

332 : 

10 

362 : 

14 

392 

: 2 

423 : 

6 

453 : 

10 

483 : 

13 

514 : 

1 

544 : 

4 

3K 

325 : 

2 

357 : 

10 

390: 

2 

422 

: 10 

455 : 

3 

4S7 : 

10 

520 : 

3 

552 : 

12 

585 : 

5 

3% 

348 : 

14 

383 : 

12 

418 : 

10 

453 

: 8 

488: 

6 

523 : 

5 

558 : 

4 

593 : 

2 

628: 

0 

3X 

373 : 

5 

410 : 

11 

448 : 

0 

4S3 

: 5 

522 : 

11 

560 : 

0 

597 : 

5 

634 : 

11 

672 : 

0 

3% 

398: 

10 

438 : 

8 

478 : 

6 

518 

: 4 

558 : 

2 

598 : 

0 

637 : 

13 

677 : 

11 

717 : 

9 

4 

? 

424 : 

10 

467 : 

2 

509: 

9 

552 

: 1 

594 : 

8 

637 : 

0 

676 : 

6 

621 : 

14 

764 : 

6 

4^ 

451 : 

11 

496 : 

14 

542 : 

2 

587 

: 5 

632 : 

7 

677 : 

10 

722: 

12 

761 : 

0 

813 : 

2 

4^ 

479 : 

8 

527 : 

7 

575 : 

6 

523 

: 6 

671 : 

5 

719 : 

4 

767 : 

3 

815 : 

2 

863 : 

2 

4% 

508 : 

3 

559 : 

0 

609 : 

13 

660 

: 10 

711 : 

6 

762 : 

4 

813 : 

0 

863 : 

15 

914 : 

11 

4 * 

537 : 

11 

591 : 

6 

645 : 

2 

698 

: 15 

752 : 

10 

806 : 

6 

860 : 

3 

913 : 

14 

967 

11 

4% 

567 : 

15 

624 : 

11 

681: 

8 

738 

: 3 

795 : 

0 

851 : 

13 

908: 

10 

965 : 

6 

1022 : 

3 

4K 

599 : 

0 

658 : 

14 

718 : 

12 

778 

• 11 

838 : 

10 

898 : 

8 

958 : 

6 

.1018 : 

5 

1078 : 

3 

4^ 

630 : 

15 

694: 

0 

757 : 

2 

820 

: 3 

883 : 

5 

946 : 

6 

1009 : 

8 

1072 : 

10 

1135 : 

12 

5 

667 : 

8 

734 : 

5 

801 : 

0 

867 

: 12 

934: 

8 

1001 : 

5 

1068 

0 

1134 : 

12 

1201: 

8 

--J 























































































































































































































































































































































































































































































TABLE VIII. 


This table exhibits at one view the weight and strength of various 
kinds of timber, as ascertained by actual experiments. 

The estimates of the weight are made when the timber is well-sea¬ 
soned and dry, but not kiln-dried. 

There may appear to be a discrepancy between the strength of tim¬ 
ber here given, and that found in the concluding remarks on Bridge 
Building; but it must be borne in mind that the calculations there 
made were on the greatest safe strain to which timber should be sub 
mitted for a long time without injury, or even impairing its elasticity, 
while the figures here given show the absolute strength, or the point 
of breakage. 


TABLE VIII. 

Weight and Strength of Timber. 


Kind of Timber. 

Weight com¬ 
pared with wa¬ 
ter — water be¬ 
ing 1090. 

No. of lbs. in 
a cubic foot. 

No. of cubic 
feet in a ton. 

| Greatest safe strain 
Greatest ten- upon a beam resting 
sile strength of upon the ends, and 
a square inch loaded in the middle ; 
in lbs. per square inch in lbs. 

White Oak, American. 

672 

42 

53 

10200 

800 

English Oak . 

930 

58 

38 

11S0O 

875 

Beech. 

S50 

42 

45 

12200 

1000 

Sycamore. 

600 

38 

59 

9600 

720 

Chestnut. 

610 

38 

59 

10650 

650 j 

Ash... 

845 

52 

43 

14100 

950 

Elm. 

670 

42 

53 

9700 

700 

Walnut. 

670 

42 

53 

8800 

675 

Poplar. 

380 

34 

66 

6900 

380 

Cedar. 

560 

33 

68 

7400 

400 

White Spruce. 

550 

34 

66 

7200 

550 

White Pine. 

590 

37 

60 

7300 

575 

Yellow Pine. 

460 

28 

80 

11800 

770 

Pitch Pine. 

660 

41 

54 

9800 

750 

. 

650 

34 

66 

9500 

675 


Strength of Timber. 


This is determined only by actual experiment, and different experi¬ 
ments give different results, according to the circumstances under which 
they are made. Timber is much stronger length-ways, with the grain, 


























































































































EXPLANATION OF TABLES. 


149 


than side-ways, across the grain; the former is called its tensile strength 
and the latter its lateral strength. In the table on page 147 we give 
both the greatest tensile and the greatest lateral strengths of different 
kinds of timber: we now propose to give some rules and data, by which 
any intelligent mechanic can make his own calculations; these data being 
based upon the strength of good white oak, to which all other kinds 
of timber may be compared. In the table it is stated that the greatest 
safe strain of oak timber, resting upon both ends and loaded in the 
middle, is 800 pounds to the square inch: that is, a stick 1 foot long 
and 1 inch square will support 800 pounds in the middle; and while 
this estimate is correct, being based upon actual experiment, yet it ap¬ 
proaches very near the breaking point, and is, no doubt, too high for 
practical estimates. We, therefore, propose 600 pounds to the square 
inch, as safe and practical, and will proceed to give our rules accordingly. 

When one end of a beam is fixed permanently in a wall and loaded 
at the other end, if the beam be made twice as long it will bear half the 
load, and if the same beam be supported at both ends it will bear eight 
times the load. If a beam be supported at both ends and loaded in the 
middle, it will bear twice the load when evenly distributed throughout 
its entire length. We are now to consider the comparative strength 
of different sticks of timber, when supported at both ends and loaded 
in the middle, when varying in length, width, and thickness. 

Rule .—Multiply the square of the perpendicular by the horizontal, 
in inches, at the center, and that product by 600; then divide by the 
length, in feet, and the quotient will be the load, in pounds, at the 
middle of the timber. 

Note .—This rule may require two corrections, although it is often 
used without them: 

1. Deduct the weight of the timber, which is estimated at 50 pounds 
to the solid foot. 

2. When the timber is more than 20 feet long, multiply by 50 less 
for every 5 feet: that is, from 20 to 25 by 550 ; from 25 to 30 by 500; 
from 30 to 35 by 450; from 35 to 40 by 400, and so on. 

See examples on next page. 


150 CARPENTRY MADE EASY. 

Examples . 

1. What is the strengths of an oak joist, 3 inches thick, 12 inches 
wide, and 16 feet long, when set on its edge? 

12 x 12 x 3 = 432 

600 

1 6)259200 

16200 

200 Weight of timber. 

16000 lbs. Ans. 

2. What is the strength of the same joist, when lying on its side ? 

3 x 3 x 12 = 108 

600 

' 16) 64800 

4050 

200 Weight of timber. 

3850 lbs. Ans. 

3. What is the strength of an oak joist, 8 inches wide, 2 inches thick, 
and 16 feet long, when standing on its edge? 

% 

8x8x2 = 128 

600 

16) 76800 

4800 

88 Weight of timber. 

4712 lbs. Ans. 

4. What is the strength of a beam 12 by 12 inches, and 20 feet 1 Eg 

12 x 12 x 12 = 1728 

600 

20) 1036800 

51840 

1000 Weight of timber. 


50840 lbs. Ans. 














EXPLANATION OF TABLES. 


151 


5. What is the strength of an oak joist, 3 inches thick, 12 inches 
wide, and 24 feet long, standing on its edge ? 

• 12 x 12 x 3 = 432 

550 

21600 

2160 

24)237600 

9900 

300 Weight of timber. 

9600 lbs. Ans. 

6. What is the strength of an oak joist, 3 inches thick, 12 inches 
wide, and 32 feet long, standing on its edge ? 

12 x 12 x 3 = 432 

500 

32)216000(6750 

192 400 Weight of timber. 

240 6350 lbs. Ans. 

224 


160 

160 

In all the above examples the strength of the timber is estimated for 
a load at the center, where the strain is the greatest. If the load is 
nearer one end, the same stick will support a greater weight, and the 
strain will be in proportion to the product of the two ends of the stick, 
measured from the point where the load rests. 

For example, if the stick is 10 feet long, and the load rests at the 
middle point, then the two ends are 5 feet each, and the strain is equal 
to 5 x 5 = 25; but if the load is 3 feet from the center, one end is 8 
feet and the other 2 feet, and the strain is equal to 8 X 2 = 16, and we 
form a proportion, thus: 

25 :16 :: 10: 6.4; so instead of dividing by 10 we divide by 6.4, and 
find the true result, and so on in similar cases. 











152 


CARPENTRY MADE EASY. 


Examples. 

1. What weight will an oak stick, 10 feet long, 10 inches wide, and 
5 inches thick, sustain, 3 feet from its center, the stick resting on its 
edge ? 

10 x 10 x 5 = 500 

600 

6.4)300000(46875 

256 175 Weight of timber. 

440 46700 lbs. A ns. 

384 


560 

512 


480 

448 


320 

320 


2. What load will an oak stick of timber sustain, 10 inches wide, 3 
inches thick, and 18 feet long, resting on its edge, the load being placed 
4 feet from the center ? 

81:65::18: 


10 x 10 x 3 = 300 bo 

600 90 

14.4)180000(12500 1Q 8 

144 175 Weight. 81)1170(14.4 

360 12325 Ans. _^L 

288 360 

720 _324 

720 36.0 


For evidence of the correctness of these rules, see Eaton's High School 
Arithmetic and Zell's Encyclopaedia, 2d Vol., p. 316. 
















The foilo?ein <y testimoziials ?v ill indicate the posi¬ 
tive value of 


“CARPENTRY MADE EASY,” 

not only for the use of apprentices, but also expe¬ 
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similar recommendations have been received . 

Sir :—Enclosed please find Five Dollars, ($5.00,) for u Carpentry made 
Easy.” 

Dear Sir, as I have examined the work thoroughly, I am more than 
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different from other works of the kind, yet so plain that any apprentice 
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Table certainly is a great saving of thought and labor. 

The work should be in the hands of all apprentices, and nine-tenths of 
the Journeymen. If it were, we would soon have good workmen, instead of 
half-hands and wood-butchers. 

Most Respectfully, 




icnautm 


Carpenter and Builder, Conshohocken, Pa. 




Another Carpenter says: 

“I am well pleased with Bell’s Carpentry. Every Carpenter whether old 
or young should have one. It is just the book for a young beginner. No 
young man should attempt to learn the carpentry trade without first buying 
“ The Art and Science of Carpentry made Easy v by Wm. E. Bell.’’ 

Millersburg, Ohio, JunG 30, 1873. 


ADDRESS, 

HOWARD CHALLEN, Publisher, 

521 MINOR STREET, Philadelphia, Pa. 

s * 

Or W. E. BELL, Ottawa, Ill. 






/*■ 


NEW EDITION—REVISED & ENLARGED. 

THE BEST WORK ISSUED FOE 

Carpenters, Builders and Mechanics. 

Forty-Four Plates and 200 Figures. 


Carpentry Made Easy, by a practical Mechanic, who fully appreciates, 
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difficult to do until they are fully explained and tested. It is designed to make the 
science and art of Carpentry clear and comparatively easy to all who require informa¬ 
tion on the subject. 

"While it will be found indispensable to the learner, there are items of information 
■worth ten times its price to any well-informed Carpenter. 

It treats on the science and art of Framing on a new and improved system, with 
specific instructions for building Balloon Frames, Barn Frames, Mill Frames, Ware¬ 
houses, Church Spires, &c. Also, a system of Bridge Building, with bills and esti¬ 
mates of cost, and valuable Tables. 

Part I—Treats on Geometry applied to Carpentry. 

Part II—Treats of the use of the Square in obtaining Bevels, Balloon Frames 
m all their departments; Roofs of all kinds, their proper pitch, length, width, dimen¬ 
sions, &c.; Church Spires, Domes, &c. 

Part III—Treats of Trestle, Arch, and Straining Beam Bridges, and general 
principles of Bridge Building. 

Part IV—Fully defines the terms and phrases used in the work, and explains 
the following tables, viz: Table 1:—Length of Common Rafters. Table 2: —Length 
of Hip Rafters. Table 3:—Octagonal Roofs. Table 4:—Length of Braces. Table 
5: —"Weight of Square Iron. Table G:—Weight of Flat Iron. Table 7:—Weight of 
Round Iron. Table 8:—Weight and Strength of Timber. 

This book is also of use to all who have barns, bridges and outhouses to build, 
and require information how to accomplish the work. 

PRICE, $5.00 PER COPY. 

ADDRESS, 

HOWARD CHALLEU, Publisher, 

521 MINOR ST., Philadelphia. 

Or W. E. BELL, Ottawa, Ill. 


O 

























































































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